Saturday, June 25, 2016

LeetCode 366 - Find Leaves of Binary Tree


http://www.cnblogs.com/grandyang/p/5616158.html
Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps until the tree is empty.
Example:
Given binary tree 
          1
         / \
        2   3
       / \     
      4   5    
Returns [4, 5, 3], [2], [1].


This problem does not actually ask you to remove leaves.
Originally, it did.
        public List<List<Integer>> findLeaves(TreeNode root) {
            List<List<Integer>> list = new ArrayList<>();
            findLeavesHelper(list, root);
            return list;
        }
        
  // return the level of root
        private int findLeavesHelper(List<List<Integer>> list, TreeNode root) {
            if (root == null) {
                return -1;
            }
            int leftLevel = findLeavesHelper(list, root.left);
            int rightLevel = findLeavesHelper(list, root.right);
            int level = Math.max(leftLevel, rightLevel) + 1;
            if (list.size() == level) {
                list.add(new ArrayList<>());
            }
            list.get(level).add(root.val);
            root.left = root.right = null;
            return level;
        }

这道题给了我们一个二叉树,让我们返回其每层的叶节点,就像剥洋葱一样,将这个二叉树一层一层剥掉,最后一个剥掉根节点。那么题目中提示说要用DFS来做,思路是这样的,每一个节点从左子节点和右子节点分开走可以得到两个深度,由于成为叶节点的条件是左右子节点都为空,所以我们取左右子节点中较大值加1为当前节点的深度值,知道了深度值就可以将节点值加入到结果res中的正确位置了,求深度的方法我们可以参见Maximum Depth of Binary Tree中求最大深度的方法
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> res;
        helper(root, res);
        return res;
    }
    int helper(TreeNode *root, vector<vector<int>> &res) {
        if (!root) return -1;
        int depth = 1 + max(helper(root->left, res), helper(root->right, res));
        if (depth >= res.size()) res.resize(depth + 1);
        res[depth].push_back(root->val);
        return depth;
    }
http://www.programcreek.com/2014/07/leetcode-find-leaves-of-binary-tree-java/
public List<List<Integer>> findLeaves(TreeNode root) {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    helper(result, root);
    return result;
}
 
// traverse the tree bottom-up recursively
private int helper(List<List<Integer>> list, TreeNode root){
    if(root==null)
        return -1;
 
    int left = helper(list, root.left);
    int right = helper(list, root.right);
    int curr = Math.max(left, right)+1;
 
    // the first time this code is reached is when curr==0,
    //since the tree is bottom-up processed.
    if(list.size()<=curr){
        list.add(new ArrayList<Integer>());
    }
 
    list.get(curr).add(root.val);
 
    return curr;
}

https://segmentfault.com/a/1190000005938045
if (list.size() == cur) {
    list.add(new ArrayList<Integer>());
}
这段代码为什么可以这么写?
因为通过跟踪递归可以发现,他的index都是连续的,没有跳跃写的情况,一般是这样:先0,再1,再2,再0,再1,再2,再3;不会出现先0,再2,再1,再4这样的情况
    public List<List<Integer>> findLeaves(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        helper(list, root);
        return list;
    }

//calculate the index of this root passed in and put it in that index, at last return where this root was put
    private int helper(List<List<Integer>> list, TreeNode root) {
        if (root == null)
            return -1;
        int left = helper(list, root.left);
        int right = helper(list, root.right);
        int cur = Math.max(left, right) + 1;
        if (list.size() == cur)
            list.add(new ArrayList<Integer>());
        list.get(cur).add(root.val);
        return cur;
    }
https://leetcode.com/discuss/110389/12-lines-simple-java-solution-using-recursion
https://discuss.leetcode.com/topic/49206/java-backtracking-o-n-time-o-n-space-no-hashing
The essential of problem is not to find the leaves, but group leaves of same level together and also to cut the tree. This is the exact role backtracking plays. The helper function returns the level which is the distance from its furthest subtree leaf to root, which helps to identify which group the root belongs to
        public List<List<Integer>> findLeaves(TreeNode root) {
            List<List<Integer>> list = new ArrayList<>();
            findLeavesHelper(list, root);
            return list;
        }
        
  // return the level of root
        private int findLeavesHelper(List<List<Integer>> list, TreeNode root) {
            if (root == null) {
                return -1;
            }
            int leftLevel = findLeavesHelper(list, root.left);
            int rightLevel = findLeavesHelper(list, root.right);
            int level = Math.max(leftLevel, rightLevel) + 1;
            if (list.size() == level) {
                list.add(new ArrayList<>());
            }
            list.get(level).add(root.val);
            root.left = root.right = null;
            return level;
        }

For this question we need to take bottom-up approach. The key is to find the height of each node. Here the definition of height is:
The height of a node is the number of edges from the node to the deepest leaf. --CMU 15-121 Binary Trees
I used a helper function to return the height of current node. According to the definition, the height of leaf is 0. h(node) = 1 + max(h(node.left), h(node.right)).
The height of a node is also the its index in the result list (res). For example, leaves, whose heights are 0, are stored in res[0]. Once we find the height of a node, we can put it directly into the result.
UPDATE:
Thanks @adrianliu0729 for pointing out that my previous code does not actually remove leaves. I added one line node.left = node.right = null; to remove visited nodes
UPDATE:
There seems to be some debate over whether we need to actually "remove" leaves from the input tree. Anyway, it is just a matter of one line code. In the actual interview, just confirm with the interviewer whether removal is required.

 Only real difference is that my extension criterion is like level == res.size(), checking equality. I think the < makes it look likeres.size() could actually be smaller than level, and I don't like giving that impression.
def findLeaves(self, root):
    def dfs(node):
        if not node:
            return -1
        i = 1 + max(dfs(node.left), dfs(node.right))
        if i == len(out):
            out.append([])
        out[i].append(node.val)
        return i
    out = []
    dfs(root)
    return out
The essential of problem is not to find the leaves, but group leaves of same level together and also to cut the tree. This is the exact role backtracking plays. The helper function returns the level which is the distance from its furthest subtree leaf to root, which helps to identify which group the root belongs to
public List<List<Integer>> findLeaves(TreeNode root) { List<List<Integer>> list = new ArrayList<>(); findLeavesHelper(list, root); return list; } // return the level of root private int findLeavesHelper(List<List<Integer>> list, TreeNode root) { if (root == null) { return -1; } int leftLevel = findLeavesHelper(list, root.left); int rightLevel = findLeavesHelper(list, root.right); int level = Math.max(leftLevel, rightLevel) + 1; if (list.size() == level) { list.add(new ArrayList<>()); } list.get(level).add(root.val); root.left = root.right = null; return level; }
https://discuss.leetcode.com/topic/49663/simple-java-dfs-solution
 public List<List<Integer>> findLeaves(TreeNode root) {
    List<List<Integer>> ans = new ArrayList<>();
    dfs(root, ans);
    return ans;
}
public int dfs(TreeNode root, List<List<Integer>> ans) {
    if (root == null) {
        return 0;
    }
    int up_cnt = Math.max(dfs(root.left, ans), dfs(root.right, ans));
    while (ans.size() <= up_cnt) {
        List<Integer> list = new ArrayList<>();
        ans.add(list);
    } 
    ans.get(up_cnt).add(root.val);
    return up_cnt + 1;
}
X. https://leetcode.com/discuss/110406/simple-java-recursive-1ms-solution
This is pretty straight forward but the general idea is to simply prune the leaves at each iteration of the while loop until the root itself is pruned. We can do this using the x = change(x) paradigm for modifying a tree. Whenever we come across a leaf node, we know we must add it to our result but then we prune it by just returning null.
Clean solution. In worst cases, it can be O(N^2) right?
private TreeNode removeLeaves(TreeNode root, List<Integer> result) { if (root == null) return null; if (root.left == null && root.right == null) { result.add(root.val); return null; } root.left = removeLeaves(root.left, result); root.right = removeLeaves(root.right, result); return root; } public List<List<Integer>> findLeaves(TreeNode root) { List<List<Integer>> results = new ArrayList<List<Integer>>(); if (root == null) return results; while (root != null) { List<Integer> leaves = new ArrayList<Integer>(); root = removeLeaves(root, leaves); results.add(leaves); } return results; }
https://discuss.leetcode.com/topic/49400/1-ms-easy-understand-java-solution
    public List<List<Integer>> findLeaves(TreeNode root) {
        
        List<List<Integer>> leavesList = new ArrayList< List<Integer>>();
        List<Integer> leaves = new ArrayList<Integer>();
        
        while(root != null) {
            if(isLeave(root, leaves)) root = null;
            leavesList.add(leaves);
            leaves = new ArrayList<Integer>();
        }
        return leavesList;
    }
    
    public boolean isLeave(TreeNode node, List<Integer> leaves) {
        
        if (node.left == null && node.right == null) {
            leaves.add(node.val);
            return true;
        }
        
        if (node.left != null) {
             if(isLeave(node.left, leaves))  node.left = null;
        }
        
        if (node.right != null) {
             if(isLeave(node.right, leaves)) node.right = null;
        }
        
        return false;
    }
下面这种DFS方法没有用计算深度的方法,而是使用了一层层剥离的方法,思路是遍历二叉树,找到叶节点,将其赋值为NULL,然后加入leaves数组中,这样一层层剥洋葱般的就可以得到最终结果了:
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> res;
        while (root) {
            vector<int> leaves;
            root = remove(root, leaves);
            res.push_back(leaves);
        }
        return res;
    }
    TreeNode* remove(TreeNode *node, vector<int> &leaves) {
        if (!node) return NULL;
        if (!node->left && !node->right) {
            leaves.push_back(node->val);
            return NULL;
        }
        node->left = remove(node->left, leaves);
        node->right = remove(node->right, leaves);
        return node;
    }
https://discuss.leetcode.com/topic/49205/java-use-map-straight
    public List<List<Integer>> findLeaves(TreeNode root) {
        HashMap<Integer, List<Integer>> m = new HashMap<>();
        List<List<Integer>> res = new LinkedList<>();
        int max = height(root, m);
        
        for(int i = 1; i<max; i++){
            if(m.containsKey(i))
                res.add(m.get(i));
        }
        return res;
    }
    
    public int height(TreeNode root, HashMap<Integer, List<Integer>> m){
        if(root == null) return 0;
        int max = Math.max(height(root.left, m), height(root.right, m)) + 1;
        if(m.containsKey(max))
            m.get(max).add(root.val); 
        else{
            LinkedList<Integer> l = new LinkedList<>();
            l.add(root.val);
            m.put(max, l);
        }
        return max+1;
    }
If you mean to make the root collectable, probably it will not happen as long as the caller still holds the reference ?

No comments:

Post a Comment

Labels

GeeksforGeeks (959) Algorithm (811) LeetCode (637) to-do (598) Review (340) Classic Algorithm (334) Classic Interview (299) Dynamic Programming (263) Google Interview (234) LeetCode - Review (229) Tree (146) POJ (137) Difficult Algorithm (136) EPI (127) Different Solutions (118) Bit Algorithms (110) Cracking Coding Interview (110) Smart Algorithm (109) Math (91) HackerRank (85) Lintcode (83) Binary Search (73) Graph Algorithm (73) Greedy Algorithm (61) Interview Corner (61) List (58) Binary Tree (56) DFS (56) Algorithm Interview (53) Advanced Data Structure (52) Codility (52) ComProGuide (52) LeetCode - Extended (47) USACO (46) Geometry Algorithm (45) BFS (43) Data Structure (42) Mathematical Algorithm (42) ACM-ICPC (41) Interval (38) Jobdu (38) Recursive Algorithm (38) Stack (38) String Algorithm (38) Binary Search Tree (37) Knapsack (37) Codeforces (36) Introduction to Algorithms (36) Matrix (36) Must Known (36) Beauty of Programming (35) Sort (35) Array (33) Trie (33) prismoskills (33) Segment Tree (32) Space Optimization (32) Union-Find (32) Backtracking (31) HDU (31) Google Code Jam (30) Permutation (30) Puzzles (30) Array O(N) (29) Data Structure Design (29) Company-Zenefits (28) Microsoft 100 - July (28) to-do-must (28) Random (27) Sliding Window (26) GeeksQuiz (25) Logic Thinking (25) hihocoder (25) High Frequency (23) Palindrome (23) Algorithm Game (22) Company - LinkedIn (22) Graph (22) Queue (22) DFS + Review (21) Hash (21) TopCoder (21) Binary Indexed Trees (20) Brain Teaser (20) CareerCup (20) Company - Twitter (20) Pre-Sort (20) Company-Facebook (19) UVA (19) Probabilities (18) Follow Up (17) Codercareer (16) Company-Uber (16) Game Theory (16) Heap (16) Shortest Path (16) String Search (16) Topological Sort (16) Tree Traversal (16) itint5 (16) Iterator (15) Merge Sort (15) O(N) (15) Difficult (14) Number (14) Number Theory (14) Post-Order Traverse (14) Priority Quieue (14) Amazon Interview (13) BST (13) Basic Algorithm (13) Bisection Method (13) Codechef (13) Majority (13) mitbbs (13) Combination (12) Computational Geometry (12) KMP (12) Long Increasing Sequence(LIS) (12) Modify Tree (12) Reconstruct Tree (12) Reservoir Sampling (12) 尺取法 (12) AOJ (11) DFS+Backtracking (11) Fast Power Algorithm (11) Graph DFS (11) LCA (11) LeetCode - DFS (11) Ordered Stack (11) Princeton (11) Tree DP (11) 挑战程序设计竞赛 (11) Binary Search - Bisection (10) Company - Microsoft (10) Company-Airbnb (10) Euclidean GCD (10) Facebook Hacker Cup (10) HackerRank Easy (10) Reverse Thinking (10) Rolling Hash (10) SPOJ (10) Theory (10) Tutorialhorizon (10) X Sum (10) Coin Change (9) Lintcode - Review (9) Mathblog (9) Max-Min Flow (9) Stack Overflow (9) Stock (9) Two Pointers (9) Book Notes (8) Bottom-Up (8) DP-Space Optimization (8) Divide and Conquer (8) Graph BFS (8) LeetCode - DP (8) LeetCode Hard (8) Prefix Sum (8) Prime (8) System Design (8) Tech-Queries (8) Time Complexity (8) Use XOR (8) 穷竭搜索 (8) Algorithm Problem List (7) DFS+BFS (7) Facebook Interview (7) Fibonacci Numbers (7) Game Nim (7) HackerRank Difficult (7) Hackerearth (7) Interval Tree (7) Linked List (7) Longest Common Subsequence(LCS) (7) Math-Divisible (7) Miscs (7) O(1) Space (7) Probability DP (7) Radix Sort (7) Simulation (7) Suffix Tree (7) Xpost (7) n00tc0d3r (7) 蓝桥杯 (7) Bucket Sort (6) Catalan Number (6) Classic Data Structure Impl (6) DFS+DP (6) DP - Tree (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Level Order Traversal (6) Manacher (6) Minimum Spanning Tree (6) One Pass (6) Programming Pearls (6) Quick Select (6) Rabin-Karp (6) Randomized Algorithms (6) Sampling (6) Schedule (6) Suffix Array (6) Threaded (6) reddit (6) AI (5) Art Of Programming-July (5) Big Data (5) Brute Force (5) Code Kata (5) Codility-lessons (5) Coding (5) Company - WMware (5) Crazyforcode (5) DFS+Cache (5) DP-Multiple Relation (5) DP-Print Solution (5) Dutch Flag (5) Fast Slow Pointers (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Inversion (5) Java (5) Kadane - Extended (5) Kadane’s Algorithm (5) Matrix Chain Multiplication (5) Microsoft Interview (5) Morris Traversal (5) Pruning (5) Quadtrees (5) Quick Partition (5) Quora (5) SPFA(Shortest Path Faster Algorithm) (5) Subarray Sum (5) Sweep Line (5) Traversal Once (5) TreeMap (5) jiuzhang (5) to-do-2 (5) 单调栈 (5) 树形DP (5) 1point3acres (4) Anagram (4) Approximate Algorithm (4) Backtracking-Include vs Exclude (4) Brute Force - Enumeration (4) Chess Game (4) Company-Amazon (4) Consistent Hash (4) Convex Hull (4) Cycle (4) DP-Include vs Exclude (4) Dijkstra (4) Distributed (4) Eulerian Cycle (4) Flood fill (4) Graph-Classic (4) HackerRank AI (4) Histogram (4) Kadane Max Sum (4) Knapsack - Mixed (4) Knapsack - Unbounded (4) Left and Right Array (4) MinMax (4) Multiple Data Structures (4) N Queens (4) Nerd Paradise (4) Parallel Algorithm (4) Practical Algorithm (4) Pre-Sum (4) Probability (4) Programcreek (4) Quick Sort (4) Spell Checker (4) Stock Maximize (4) Subsets (4) Sudoku (4) Symbol Table (4) TreeSet (4) Triangle (4) Water Jug (4) Word Ladder (4) algnotes (4) fgdsb (4) 最大化最小值 (4) A Star (3) Abbreviation (3) Algorithm - Brain Teaser (3) Algorithm Design (3) Anagrams (3) B Tree (3) Big Data Algorithm (3) Binary Search - Smart (3) Caterpillar Method (3) Coins (3) Company - Groupon (3) Company - Indeed (3) Cumulative Sum (3) DP-Fill by Length (3) DP-Two Variables (3) Dedup (3) Dequeue (3) Dropbox (3) Easy (3) Edit Distance (3) Expression (3) Finite Automata (3) Forward && Backward Scan (3) Github (3) GoLang (3) Include vs Exclude (3) Joseph (3) Jump Game (3) Knapsack-多重背包 (3) LeetCode - Bit (3) LeetCode - TODO (3) Linked List Merge Sort (3) LogN (3) Master Theorem (3) Maze (3) Min Cost Flow (3) Minesweeper (3) Missing Numbers (3) NP Hard (3) Online Algorithm (3) Pascal's Triangle (3) Pattern Match (3) Project Euler (3) Rectangle (3) Scala (3) SegmentFault (3) Stack - Smart (3) State Machine (3) Streaming Algorithm (3) Subset Sum (3) Subtree (3) Transform Tree (3) Two Pointers Window (3) Warshall Floyd (3) With Random Pointer (3) Word Search (3) bookkeeping (3) codebytes (3) Activity Selection Problem (2) Advanced Algorithm (2) AnAlgorithmADay (2) Application of Algorithm (2) Array Merge (2) BOJ (2) BT - Path Sum (2) Balanced Binary Search Tree (2) Bellman Ford (2) Binomial Coefficient (2) Bit Mask (2) Bit-Difficult (2) Bloom Filter (2) Book Coding Interview (2) Branch and Bound Method (2) Clock (2) Codesays (2) Company - Baidu (2) Complete Binary Tree (2) DFS+BFS, Flood Fill (2) DP - DFS (2) DP-3D Table (2) DP-Classical (2) DP-Output Solution (2) DP-Slide Window Gap (2) DP-i-k-j (2) DP-树形 (2) Distributed Algorithms (2) Divide and Conqure (2) Doubly Linked List (2) GoHired (2) Graham Scan (2) Graph - Bipartite (2) Graph BFS+DFS (2) Graph Coloring (2) Graph-Cut Vertices (2) Hamiltonian Cycle (2) Huffman Tree (2) In-order Traverse (2) Include or Exclude Last Element (2) Information Retrieval (2) Interview - Linkedin (2) Invariant (2) Islands (2) Knuth Shuffle (2) LeetCode - Recursive (2) Linked Interview (2) Linked List Sort (2) Longest SubArray (2) Lucene-Solr (2) MST (2) MST-Kruskal (2) Math-Remainder Queue (2) Matrix Power (2) Minimum Vertex Cover (2) Negative All Values (2) Number Each Digit (2) Numerical Method (2) Object Design (2) Order Statistic Tree (2) Palindromic (2) Parentheses (2) Parser (2) Peak (2) Programming (2) Range Minimum Query (2) Reuse Forward Backward (2) Robot (2) Rosettacode (2) Scan from right (2) Search (2) Shuffle (2) Sieve of Eratosthenes (2) SimHash (2) Simple Algorithm (2) Skyline (2) Spatial Index (2) Stream (2) Strongly Connected Components (2) Summary (2) TV (2) Tile (2) Traversal From End (2) Tree Sum (2) Tree Traversal Return Multiple Values (2) Word Break (2) Word Graph (2) Word Trie (2) Young Tableau (2) 剑指Offer (2) 数位DP (2) 1-X (1) 51Nod (1) Akka (1) Algorithm - How To (1) Algorithm - New (1) Algorithm Series (1) Algorithms Part I (1) Analysis of Algorithm (1) Array-Element Index Negative (1) Array-Rearrange (1) Auxiliary Array (1) Auxiliary Array: Inc&Dec (1) BACK (1) BK-Tree (1) BZOJ (1) Basic (1) Bayes (1) Beauty of Math (1) Big Integer (1) Big Number (1) Binary (1) Binary Tree Variant (1) Bipartite (1) Bit-Missing Number (1) BitMap (1) BitMap index (1) BitSet (1) Bug Free Code (1) BuildIt (1) C/C++ (1) CC Interview (1) Cache (1) Calculate Height at Same Recusrion (1) Cartesian tree (1) Check Tree Property (1) Chinese (1) Circular Buffer (1) Code Quality (1) Codesolutiony (1) Company - Alibaba (1) Company - Palantir (1) Company - WalmartLabs (1) Company-Apple (1) Company-Epic (1) Company-Salesforce (1) Company-Snapchat (1) Company-Yelp (1) Compression Algorithm (1) Concurrency (1) Convert BST to DLL (1) Convert DLL to BST (1) Custom Sort (1) Cyclic Replacement (1) DFS-Matrix (1) DP - Probability (1) DP Fill Diagonal First (1) DP-Difficult (1) DP-End with 0 or 1 (1) DP-Fill Diagonal First (1) DP-Graph (1) DP-Left and Right Array (1) DP-MaxMin (1) DP-Memoization (1) DP-Node All Possibilities (1) DP-Optimization (1) DP-Preserve Previous Value (1) DP-Print All Solution (1) Database (1) Detect Negative Cycle (1) Directed Graph (1) Do Two Things at Same Recusrion (1) Domino (1) Dr Dobb's (1) Duplicate (1) Equal probability (1) External Sort (1) FST (1) Failure Function (1) Fraction (1) Front End Pointers (1) Funny (1) Fuzzy String Search (1) Game (1) Generating Function (1) Generation (1) Genetic algorithm (1) GeoHash (1) Geometry - Orientation (1) Google APAC (1) Graph But No Graph (1) Graph Transpose (1) Graph Traversal (1) Graph-Coloring (1) Graph-Longest Path (1) Gray Code (1) HOJ (1) Hanoi (1) Hard Algorithm (1) How Hash (1) How to Test (1) Improve It (1) In Place (1) Inorder-Reverse Inorder Traverse Simultaneously (1) Interpolation search (1) Interview (1) Interview - Easy (1) Interview - Facebook (1) Isomorphic (1) JDK8 (1) K Dimensional Tree (1) Knapsack - Fractional (1) Knapsack - ZeroOnePack (1) Knight (1) Kosaraju’s algorithm (1) Kruskal (1) Kruskal MST (1) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Detail (1) LeetCode - Related (1) LeetCode Diffcult (1) Linked List Reverse (1) Linkedin (1) Linkedin Interview (1) Local MinMax (1) Logic Pattern (1) Longest Common Subsequence (1) Longest Common Substring (1) Longest Prefix Suffix(LPS) (1) Manhattan Distance (1) Map && Reverse Map (1) Math - Induction (1) Math-Multiply (1) Math-Sum Of Digits (1) Matrix - O(N+M) (1) Matrix BFS (1) Matrix Graph (1) Matrix Search (1) Matrix+DP (1) Matrix-Rotate (1) Max Min So Far (1) Median (1) Memory-Efficient (1) MinHash (1) MinMax Heap (1) Monotone Queue (1) Monto Carlo (1) Multi-Reverse (1) Multiple DFS (1) Multiple Tasks (1) Next Successor (1) Offline Algorithm (1) PAT (1) Partition (1) Path Finding (1) Patience Sort (1) Persistent (1) Pigeon Hole Principle (1) Power Set (1) Pratical Algorithm (1) Probabilistic Data Structure (1) Proof (1) Python (1) Queue & Stack (1) RSA (1) Ranking (1) Rddles (1) ReHash (1) Realtime (1) Recurrence Relation (1) Recursive DFS (1) Recursive to Iterative (1) Red-Black Tree (1) Region (1) Regular Expression (1) Resources (1) Reverse Inorder Traversal (1) Robin (1) Selection (1) Self Balancing BST (1) Similarity (1) Sort && Binary Search (1) String Algorithm. Symbol Table (1) String DP (1) String Distance (1) SubMatrix (1) Subsequence (1) System of Difference Constraints(差分约束系统) (1) TSP (1) Ternary Search Tree (1) Test (1) Thread (1) TimSort (1) Top-Down (1) Tournament (1) Tournament Tree (1) Transform Tree in Place (1) Tree Diameter (1) Tree Rotate (1) Trie + DFS (1) Trie and Heap (1) Trie vs Hash (1) Trie vs HashMap (1) Triplet (1) Two Data Structures (1) Two Stacks (1) USACO - Classical (1) USACO - Problems (1) UyHiP (1) Valid Tree (1) Vector (1) Wiggle Sort (1) Wikipedia (1) Yahoo Interview (1) ZOJ (1) baozitraining (1) codevs (1) cos126 (1) javabeat (1) jum (1) namic Programming (1) sqrt(N) (1) 两次dijkstra (1) 九度 (1) 二进制枚举 (1) 夹逼法 (1) 归一化 (1) 折半枚举 (1) 枚举 (1) 状态压缩DP (1) 男人八题 (1) 英雄会 (1) 逆向思维 (1)

Popular Posts