Wednesday, November 2, 2016

LeetCode 449 - Serialize and Deserialize BST


TODO - LeetCode 450 - Delete Node in a BST
https://leetcode.com/problems/serialize-and-deserialize-bst/
Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a binary search tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary search tree can be serialized to a string and this string can be deserialized to the original tree structure.
The encoded string should be as compact as possible.
Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.
https://discuss.leetcode.com/topic/66832/java-o-n-recursive-dfs-without-null-changed-from-serialize-and-deserialize-bt
Thanks to this post, I realize that I can make use of lower and upper bound.
public String serialize(TreeNode root) { // preorder
    StringBuilder sb = new StringBuilder();
    serializedfs(root, sb);
    return sb.toString();
}

private void serializedfs(TreeNode root, StringBuilder sb){
    if(root == null) return; // no "null "
    sb.append(root.val).append(" ");
    serializedfs(root.left, sb);
    serializedfs(root.right, sb);
}

// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
    if(data.length() == 0) return null;
    String[] list = data.split(" ");
    TreeNode dummy = new TreeNode(0);
    deserializedfs(list, 0, dummy, true, Integer.MIN_VALUE, Integer.MAX_VALUE);
    return dummy.left;
}

private int deserializedfs(String[] list, int pos, TreeNode par, boolean isleft, 
                                                    int lower, int upper){
    if(pos >= list.length) return pos;

    int val = Integer.valueOf(list[pos]);
    if(val < lower || val > upper) return pos-1; // have not used this pos, so minus one
    TreeNode cur = new TreeNode(val);
    
    if(isleft) par.left = cur;
    else       par.right = cur;

    pos = deserializedfs(list, ++pos, cur, true, lower, val);
    pos = deserializedfs(list, ++pos, cur, false, val, upper);
    return pos;
}
http://www.cnblogs.com/grandyang/p/4913869.html
先序遍历的递归解法,非常的简单易懂,我们需要接入输入和输出字符串流istringstream和ostringstream,对于序列化,我们从根节点开始,如果节点存在,则将值存入输出字符串流,然后分别对其左右子节点递归调用序列化函数即可。对于去序列化,我们先读入第一个字符,以此生成一个根节点,然后再对根节点的左右子节点递归调用去序列化函数即可
    string serialize(TreeNode* root) {
        ostringstream out;
        serialize(root, out);
        return out.str();
    }
    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        istringstream in(data);
        return deserialize(in);
    }
private:
    void serialize(TreeNode *root, ostringstream &out) {
        if (root) {
            out << root->val << ' ';
            serialize(root->left, out);
            serialize(root->right, out);
        } else {
            out << "# ";
        }
    }
    TreeNode* deserialize(istringstream &in) {
        string val;
        in >> val;
        if (val == "#") return nullptr;
        TreeNode *root = new TreeNode(stoi(val));
        root->left = deserialize(in);
        root->right = deserialize(in);
        return root;
    }

http://bookshadow.com/weblog/2016/11/01/leetcode-serialize-and-deserialize-bst/
先序遍历(Preorder Traversal)
根据二叉搜索树(BST)的性质,左孩子 < 根节点 < 右孩子,因此可以通过先序遍历的结果唯一确定一棵原始二叉树。
序列化(Serialization):
先序遍历原始二叉树,输出逗号分隔值字符串。
反序列化(Deserialization):
利用栈(Stack)数据结构,节点栈nstack保存重建二叉树过程中的节点,最大值栈rstack保存当前节点的右孩子允许的最大值。

遍历序列化串,记当前数值为val,新增树节点node = TreeNode(val);记nstack的栈顶元素为ntop(nstack[-1])

若val < ntop,则val为ntop的左孩子,令ntop.left = node,并将node压入nstack

否则,val应为右孩子,但其父节点并不一定为ntop:

    记rstack的栈顶元素为rtop,当val > rtop时,执行循环:
    
        重复弹出nstack,直到ntop不是右孩子为止(nstack[-1] > nstack[-2]条件不成立)
        
        再次弹出nstack, 并弹出rstack

    上述过程执行完毕后,令ntop.right = node,并将node压入nstack
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Codec: def serialize(self, root): """Encodes a tree to a single string. :type root: TreeNode :rtype: str """ if not root: return [] left = self.serialize(root.left) right = self.serialize(root.right) ans = str(root.val) if left: ans += ',' + left if right: ans += ',' + right return ans def deserialize(self, data): """Decodes your encoded data to tree. :type data: str :rtype: TreeNode """ if not data: return None nstack, rstack = [], [0x7FFFFFFF] for val in map(int, data.split(',')): node = TreeNode(val) if nstack: if val < nstack[-1].val: nstack[-1].left = node rstack.append(nstack[-1].val) else: while val > rstack[-1]: while nstack[-1].val > nstack[-2].val: nstack.pop() rstack.pop() nstack.pop() nstack[-1].right = node nstack.append(node) return nstack[0] # Your Codec object will be instantiated and called as such: # codec = Codec() # codec.deserialize(codec.serialize(root))
X. BFS, Queue
http://www.cnblogs.com/charlesblc/p/6019644.html
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

class Codec {
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        if (root == null) {
            return "";
        }

        // LinkedList实现了Queue接口, ArrayList没有实现
        Queue<TreeNode> qe= new LinkedList<>();
        qe.offer(root);
        sb.append(root.val+",");
        while (!qe.isEmpty()) {
            TreeNode tn = qe.poll();
            if (tn.left != null) {
                sb.append(tn.left.val+",");
                qe.offer(tn.left);
            }
            else {
                sb.append(",");
            }
            if (tn.right != null) {
                sb.append(tn.right.val+",");
                qe.offer(tn.right);
            }
            else {
                sb.append(",");
            }
        }
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data.equals("")) {
            return null;
        }

        String[] strs = data.split(",");
        Queue<TreeNode> qe = new LinkedList<>();

        if (strs.length < 1 || strs[0].equals("")) {
            return null;
        }

        TreeNode root = new TreeNode(Integer.valueOf(strs[0]));
        qe.offer(root);
        int i = 1;
        while (!qe.isEmpty()) {
            TreeNode tn = qe.poll();

            if (strs.length > i && !strs[i].equals("")) {
                TreeNode left = new TreeNode(Integer.valueOf(strs[i]));
                tn.left = left;
                qe.offer(left);
            }
            i++;
            if (strs.length > i && !strs[i].equals("")) {
                TreeNode right = new TreeNode(Integer.valueOf(strs[i]));
                tn.right = right;
                qe.offer(right);
            }
            i++;
        }
        return root;
    }
}

另一种方法是层序遍历的非递归解法,这种方法略微复杂一些,我们需要借助queue来做,本质是BFS算法,也不是很难理解,就是BFS算法的常规套路稍作修改即可
    string serialize(TreeNode* root) {
        ostringstream out;
        queue<TreeNode*> q;
        if (root) q.push(root);
        while (!q.empty()) {
            TreeNode *t = q.front(); q.pop();
            if (t) {
                out << t->val << ' ';
                q.push(t->left);
                q.push(t->right);
            } else {
                out << "# ";
            }
        }
        return out.str();
    }
    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        if (data.empty()) return nullptr;
        istringstream in(data);
        queue<TreeNode*> q;
        string val;
        in >> val;
        TreeNode *res = new TreeNode(stoi(val)), *cur = res;
        q.push(cur);
        while (!q.empty()) {
            TreeNode *t = q.front(); q.pop();
            if (!(in >> val)) break;
            if (val != "#") {
                cur = new TreeNode(stoi(val));
                q.push(cur);
                t->left = cur;
            }
            if (!(in >> val)) break;
            if (val != "#") {
                cur = new TreeNode(stoi(val));
                q.push(cur);
                t->right = cur;
            }
        }
        return res;
    }

https://discuss.leetcode.com/topic/66760/java-bfs-solution-easy-to-understand-serialize-deserialize-runtime-o-n
Each tree node can be represented by "val/num/", where val is the value of the node and num indicate his children situation (num == 3 meaning having two children, num == 2 meaning having only left child, num == 1 meaning having only right child, num == 0 meaning having no child).
The time complexity for both serialize and deserialize is O(n), where n is the number of nodes in BST. The trade-off here is that I use an extra char "num" as in val/num/.
    // Encodes a tree to a single string.
 public String serialize(TreeNode root) {
        if (root == null) return "";
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        StringBuilder sb = new StringBuilder();
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            int num = 0;
            if (node.left != null && node.right != null) {
                // 3 indicates having both left and right child
                num = 3;
                queue.offer(node.left);
                queue.offer(node.right);
            } else if (node.left != null) {
                // 2 indicates having left child
                num = 2;
                queue.offer(node.left);
            } else if (node.right != null) {
                // 1 indicates having right child
                num = 1;
                queue.offer(node.right);
            } // 0 indicates having no child
            
            sb.append(node.val).append("/").append(num).append("/");
        }
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || data.length() < 4) return null;
        char[] text = data.toCharArray();
        int i = 0;
        TreeNode root = new TreeNode(-1);
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            // Set node value
            i = readNode(text, i, node);
            // Read node's child number
            int num = 0;
            while (text[i] != '/') {
                num = num*10 + Character.getNumericValue(text[i]);
                i++;
            }
            i++;
            if (num == 3) {
                addLeftNode(node, queue);
                addRightNode(node, queue);
            } else if (num == 2) {
                addLeftNode(node, queue);
            } else if (num == 1) {
                addRightNode(node, queue);
            }
        }
        return root;
    }
    
    private int readNode(char[] text, int i, TreeNode node) {
        int val = 0;
        while (text[i] != '/') {
            val = val*10 + Character.getNumericValue(text[i]);
            i++;
        }
        node.val = val;
        return i+1;
    }
    
    private void addLeftNode(TreeNode parent, Queue<TreeNode> queue) {
        TreeNode node = new TreeNode(-1);
        parent.left = node;
        queue.offer(node);
    }
    
    private void addRightNode(TreeNode parent, Queue<TreeNode> queue) {
        TreeNode node = new TreeNode(-1);
        parent.right = node;
        queue.offer(node);
    }
}

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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)

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