Binary Tree Zigzag Level Order Traversal -- leetcode

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]
]

基本思路：

1. 使用广度优先搜索

2. 交替使用两个栈。一个表示当前层，一个表示下一层的节点。

3. 使用一个标志变量，表示该层中，先入左节点，还是先入右结点。

在leetcode上实际执行时间为4ms。

/**
 * Definition for a binary tree node.
 * 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) {
        vector<vector<int> > ans;
        if (!root) return ans;
        stack<TreeNode *> *next = new stack<TreeNode *>();
        stack<TreeNode *> *curr = new stack<TreeNode *>();
        curr->push(root);
        bool flag = true;
        while (!curr->empty()) {
            ans.push_back(vector<int>());
            ans.back().reserve(curr->size());
            while (!curr->empty()) {
                root = curr->top();
                curr->pop();
                ans.back().push_back(root->val);
                if (flag) {
                    if (root->left) next->push(root->left);
                    if (root->right) next->push(root->right);
                }
                else {
                    if (root->right) next->push(root->right);
                    if (root->left) next->push(root->left);
                }
            }
            swap(next, curr);
            flag = !flag;
        }
        
        delete next;
        delete curr;
        return ans;
    }
};

Binary Tree Zigzag Level Order Traversal -- leetcode

原文：http://blog.csdn.net/elton_xiao/article/details/45537385