Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

[  [3],  [20,9],  [15,7]]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

用两个栈实现

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function                vector<vector<int> > ret;        stack<TreeNode*> currentLevel;        stack<TreeNode*> nextLevel;        int flag = 1;                if(root == NULL) return ret;                currentLevel.push(root);                stack<TreeNode*> stack ;                while(1){            vector<int> vec;            while(!currentLevel.empty()){                TreeNode* current = currentLevel.top();                currentLevel.pop();                vec.push_back(current->val);                if(flag){                    if(current->left != NULL) nextLevel.push(current->left);                    if(current->right != NULL) nextLevel.push(current->right);                }else{                    if(current->right != NULL) nextLevel.push(current->right);                    if(current->left != NULL) nextLevel.push(current->left);                }            }            flag = 1- flag;            ret.push_back(vec);            if(nextLevel.empty()) break;            stack = currentLevel;            currentLevel = nextLevel;            nextLevel = stack;            }                return ret;            }};

28 milli secs