LeetCode 530 - Minimum Absolute Difference in BST


https://leetcode.com/problems/minimum-absolute-difference-in-bst
Input:

   1
    \
     3
    /
   2

Output:
1

Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).
Note: There are at least two nodes in this BST.

X. Inorder travsere
https://discuss.leetcode.com/topic/80796/java-o-n-time-inorder-traversal-solution
Since this is a BST, the inorder traversal of its nodes results in a sorted list of values. Thus, the minimum absolute difference must occur in any adjacently traversed nodes. I use the global variable "prev" to keep track of each node's inorder predecessor.
    int minDiff = Integer.MAX_VALUE;
    TreeNode prev;
    
    public int getMinimumDifference(TreeNode root) {
        inorder(root);
        return minDiff;
    }
    
    public void inorder(TreeNode root) {
        if (root == null) return;
        inorder(root.left);
        if (prev != null) minDiff = Math.min(minDiff, root.val - prev.val);
        prev = root;
        inorder(root.right);
    }
Indeed, hijacking a mutable array to emulate pass-by-reference reeks of a recovering C++ programmer (guilty as charged ;) )
class WrapInt{
    int value;
}
    public int getMinimumDifference(TreeNode root) {
     List<Integer> prev = new ArrayList<>(); // contains at most 1 value
     int[] min = new int[]{Integer.MAX_VALUE};
     inorder(root, prev, min);
     return min[0];
    }
    
    private void inorder(TreeNode root, List<Integer> prev, int[] min) {
     if (root == null) return;
     
     inorder(root.left, prev, min);
     if (prev.isEmpty()) {
      prev.add(root.val);
     } else {
      min[0] = Math.min(min[0], Math.abs(root.val - prev.get(0)));
      prev.set(0, root.val);
     }
     inorder(root.right, prev, min);     
    }
https://discuss.leetcode.com/topic/80823/twof-solutions-in-order-traversal-and-a-more-general-way-using-treeset
Solution 1 - In-Order traverse, time complexity O(N), space complexity O(1).
Could you explain using Integer.MAX_VALUE;
int min = Integer.MAX_VALUE;
That's just for initializing the return value. Because when you call
min = Math.min(min, Math.abs(root.val - set.floor(root.val)));
the first time, you want min to be whatever the right part is, which can be achieved by setting min to have the maximum possible value.
public class Solution {
    int min = Integer.MAX_VALUE;//
    Integer prev = null;
    
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return min;
        
        getMinimumDifference(root.left);
        
        if (prev != null) {
            min = Math.min(min, root.val - prev);
        }
        prev = root.val;
        
        getMinimumDifference(root.right);
        
        return min;
    }
    
}
What if it is not a BST? (Follow up of the problem) The idea is to put values in a TreeSet and then every time we can use O(lgN) time to lookup for the nearest values.

Solution 2 - Pre-Order traverse, time complexity O(NlgN), space complexity O(N).
    TreeSet<Integer> set = new TreeSet<>();
    int min = Integer.MAX_VALUE;
    
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return min;
        
        if (!set.isEmpty()) {
            if (set.floor(root.val) != null) {
                min = Math.min(min, Math.abs(root.val - set.floor(root.val)));
            }
            if (set.ceiling(root.val) != null) {
                min = Math.min(min, Math.abs(root.val - set.ceiling(root.val)));
            }
        }
        
        set.add(root.val);
        
        getMinimumDifference(root.left);
        getMinimumDifference(root.right);
        
        return min;
    }
中序遍历BST得到的序列是有序序列。因此创建一个vector存储中序遍历的每一个节点就得到一个有序序列,然后遍历这个有序序列,在遍历过程中求出两两之差并更新最小的绝对值之差就能得到结果。
http://www.liuchuo.net/archives/3714
    int getMinimumDifference(TreeNode* root) {
        inOrder(root);
        int result = INT_MAX;
        for (int i = 1; i < tree.size(); i++)
            result = min(result, tree[i] - tree[i-1]);
        return result;
    }
private:
    vector<int> tree;
    void inOrder(TreeNode* root) {
        if (root->left != NULL) inOrder(root->left);
        tree.push_back(root->val);
        if (root->right != NULL) inOrder(root->right);
    }
};

X. [lower, upper bound]
https://discuss.leetcode.com/topic/80916/java-no-in-order-traverse-solution-just-pass-upper-bound-and-lower-bound
It is just sort of property of the BST. Since the value of all the nodes of the left subtree is smaller than the current node, the value of the current node is the upper bound of the all the nodes in the left subtree. And for the right subtree, the value of the current node is the lower bound. You can draw a BST and have a look. If you draw a vertical line for 2 nodes, you will find all the nodes between those two lines are bounded by the value of the nodes which you draw the vertical lines.
Make use of the property of BST that value of nodes is bounded by their "previous" and "next" node.
Edit: Thanks to Stefan pointing out a small bug. Previous code would fail when testing [2147483647, 2147483646].
Now Long.MAX_VALUE/MIN_VALUE is used to mark the INF.
long minDiff = Long.MAX_VALUE;
public int getMinimumDifference(TreeNode root) {
    helper(root,Long.MIN_VALUE,Long.MAX_VALUE);
    return (int)minDiff;
}
private void helper(TreeNode curr, long lb, long rb){
    if(curr==null) return;
    if(lb!=Long.MIN_VALUE){
        minDiff = Math.min(minDiff,curr.val - lb);
    }
    if(rb!=Long.MAX_VALUE){
    minDiff = Math.min(minDiff,rb - curr.val);
    }
    helper(curr.left,lb,curr.val);
    helper(curr.right,curr.val,rb);
}
def getMinimumDifference(self, root):
    def mindiff(root, lo, hi):
        if not root:
            return float('inf')
        return min(root.val - lo,
                   hi - root.val,
                   mindiff(root.left, lo, root.val),
                   mindiff(root.right, root.val, hi))
    return mindiff(root, float('-inf'), float('inf'))
X.
https://leetcode.com/problems/minimum-absolute-difference-in-bst/discuss/99905/Two-Solutions-in-order-traversal-and-a-more-general-way-using-TreeSet
What if it is not a BST? (Follow up of the problem) The idea is to put values in a TreeSet and then every time we can use O(lgN) time to lookup for the nearest values.
Solution 2 - Pre-Order traverse, time complexity O(NlgN), space complexity O(N).
public class Solution {
    TreeSet<Integer> set = new TreeSet<>();
    int min = Integer.MAX_VALUE;
    
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return min;
        
        if (!set.isEmpty()) {
            if (set.floor(root.val) != null) {
                min = Math.min(min, root.val - set.floor(root.val));
            }
            if (set.ceiling(root.val) != null) {
                min = Math.min(min, set.ceiling(root.val) - root.val);
            }
        }
        
        set.add(root.val);
        
        getMinimumDifference(root.left);
        getMinimumDifference(root.right);
        
        return min;
    }
}

Labels

LeetCode (1432) GeeksforGeeks (1122) LeetCode - Review (1067) Review (882) Algorithm (668) to-do (609) Classic Algorithm (270) Google Interview (237) Classic Interview (222) Dynamic Programming (220) DP (186) Bit Algorithms (145) POJ (141) Math (137) Tree (132) LeetCode - Phone (129) EPI (122) Cracking Coding Interview (119) DFS (115) Difficult Algorithm (115) Lintcode (115) Different Solutions (110) Smart Algorithm (104) Binary Search (96) BFS (91) HackerRank (90) Binary Tree (86) Hard (79) Two Pointers (78) Stack (76) Company-Facebook (75) BST (72) Graph Algorithm (72) Time Complexity (69) Greedy Algorithm (68) Interval (63) Company - Google (62) Geometry Algorithm (61) Interview Corner (61) LeetCode - Extended (61) Union-Find (60) Trie (58) Advanced Data Structure (56) List (56) Priority Queue (53) Codility (52) ComProGuide (50) LeetCode Hard (50) Matrix (50) Bisection (48) Segment Tree (48) Sliding Window (48) USACO (46) Space Optimization (45) Company-Airbnb (41) Greedy (41) Mathematical Algorithm (41) Tree - Post-Order (41) ACM-ICPC (40) Algorithm Interview (40) Data Structure Design (40) Graph (40) Backtracking (39) Data Structure (39) Jobdu (39) Random (39) Codeforces (38) Knapsack (38) LeetCode - DP (38) Recursive Algorithm (38) String Algorithm (38) TopCoder (38) Sort (37) Introduction to Algorithms (36) Pre-Sort (36) Beauty of Programming (35) Must Known (34) Binary Search Tree (33) Follow Up (33) prismoskills (33) Palindrome (32) Permutation (31) Array (30) Google Code Jam (30) HDU (30) Array O(N) (29) Logic Thinking (29) Monotonic Stack (29) Puzzles (29) Code - Detail (27) Company-Zenefits (27) Microsoft 100 - July (27) Queue (27) Binary Indexed Trees (26) TreeMap (26) to-do-must (26) 1point3acres (25) GeeksQuiz (25) Merge Sort (25) Reverse Thinking (25) hihocoder (25) Company - LinkedIn (24) Hash (24) High Frequency (24) Summary (24) Divide and Conquer (23) Proof (23) Game Theory (22) Topological Sort (22) Lintcode - Review (21) Tree - Modification (21) Algorithm Game (20) CareerCup (20) Company - Twitter (20) DFS + Review (20) DP - Relation (20) Brain Teaser (19) DP - Tree (19) Left and Right Array (19) O(N) (19) Sweep Line (19) UVA (19) DP - Bit Masking (18) LeetCode - Thinking (18) KMP (17) LeetCode - TODO (17) Probabilities (17) Simulation (17) String Search (17) Codercareer (16) Company-Uber (16) Iterator (16) Number (16) O(1) Space (16) Shortest Path (16) itint5 (16) DFS+Cache (15) Dijkstra (15) Euclidean GCD (15) Heap (15) LeetCode - Hard (15) Majority (15) Number Theory (15) Rolling Hash (15) Tree Traversal (15) Brute Force (14) Bucket Sort (14) DP - Knapsack (14) DP - Probability (14) Difficult (14) Fast Power Algorithm (14) Pattern (14) Prefix Sum (14) TreeSet (14) Algorithm Videos (13) Amazon Interview (13) Basic Algorithm (13) Codechef (13) Combination (13) Computational Geometry (13) DP - Digit (13) LCA (13) LeetCode - DFS (13) Linked List (13) Long Increasing Sequence(LIS) (13) Math-Divisible (13) Reservoir Sampling (13) mitbbs (13) Algorithm - How To (12) Company - Microsoft (12) DP - Interval (12) DP - Multiple Relation (12) DP - Relation Optimization (12) LeetCode - Classic (12) Level Order Traversal (12) Prime (12) Pruning (12) Reconstruct Tree (12) Thinking (12) X Sum (12) AOJ (11) Bit Mask (11) Company-Snapchat (11) DP - Space Optimization (11) Dequeue (11) Graph DFS (11) MinMax (11) Miscs (11) Princeton (11) Quick Sort (11) Stack - Tree (11) 尺取法 (11) 挑战程序设计竞赛 (11) Coin Change (10) DFS+Backtracking (10) Facebook Hacker Cup (10) Fast Slow Pointers (10) HackerRank Easy (10) Interval Tree (10) Limited Range (10) Matrix - Traverse (10) Monotone Queue (10) SPOJ (10) Starting Point (10) States (10) Stock (10) Theory (10) Tutorialhorizon (10) Kadane - Extended (9) Mathblog (9) Max-Min Flow (9) Maze (9) Median (9) O(32N) (9) Quick Select (9) Stack Overflow (9) System Design (9) Tree - Conversion (9) Use XOR (9) Book Notes (8) Company-Amazon (8) DFS+BFS (8) DP - States (8) Expression (8) Longest Common Subsequence(LCS) (8) One Pass (8) Quadtrees (8) Traversal Once (8) Trie - Suffix (8) 穷竭搜索 (8) Algorithm Problem List (7) All Sub (7) Catalan Number (7) Cycle (7) DP - Cases (7) Facebook Interview (7) Fibonacci Numbers (7) Flood fill (7) Game Nim (7) Graph BFS (7) HackerRank Difficult (7) Hackerearth (7) Inversion (7) Kadane’s Algorithm (7) Manacher (7) Morris Traversal (7) Multiple Data Structures (7) Normalized Key (7) O(XN) (7) Radix Sort (7) Recursion (7) Sampling (7) Suffix Array (7) Tech-Queries (7) Tree - Serialization (7) Tree DP (7) Trie - Bit (7) 蓝桥杯 (7) Algorithm - Brain Teaser (6) BFS - Priority Queue (6) BFS - Unusual (6) Classic Data Structure Impl (6) DP - 2D (6) DP - Monotone Queue (6) DP - Unusual (6) DP-Space Optimization (6) Dutch Flag (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Local MinMax (6) MST (6) Minimum Spanning Tree (6) Number - Reach (6) Parentheses (6) Pre-Sum (6) Probability (6) Programming Pearls (6) Rabin-Karp (6) Reverse (6) Scan from right (6) Schedule (6) Stream (6) Subset Sum (6) TSP (6) Xpost (6) n00tc0d3r (6) reddit (6) AI (5) Abbreviation (5) Anagram (5) Art Of Programming-July (5) Assumption (5) Bellman Ford (5) Big Data (5) Code - Solid (5) Code Kata (5) Codility-lessons (5) Coding (5) Company - WMware (5) Convex Hull (5) Crazyforcode (5) DFS - Multiple (5) DFS+DP (5) DP - Multi-Dimension (5) DP-Multiple Relation (5) Eulerian Cycle (5) Graph - Unusual (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Java (5) LogN (5) Manhattan Distance (5) Matrix Chain Multiplication (5) N Queens (5) Pre-Sort: Index (5) Quick Partition (5) Quora (5) Randomized Algorithms (5) Resources (5) Robot (5) SPFA(Shortest Path Faster Algorithm) (5) Shuffle (5) Sieve of Eratosthenes (5) Strongly Connected Components (5) Subarray Sum (5) Sudoku (5) Suffix Tree (5) Swap (5) Threaded (5) Tree - Creation (5) Warshall Floyd (5) Word Search (5) jiuzhang (5)

Popular Posts