【Leetcode】【Medium】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]
]

解题思路：

按层遍历，建立两个栈，一个保存当前结点，一个按照Z型保存下一层结点。

使用一个bool变量来标注每次读取子节点的方向；

代码：

 /**
  * 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> > ret;
         stack<TreeNode*> cur_layer;
         stack<TreeNode*> next_layer;
         cur_layer.push(root);
         bool dirt = false;

         if (!root)
             return ret;

         while (!cur_layer.empty()) {
             vector<int> layer_val;
             while (!cur_layer.empty()) {
                 TreeNode* node = cur_layer.top();
                 cur_layer.pop();
                 layer_val.push_back(node->val);
                 if (dirt) {
                     if (node->right)
                         next_layer.push(node->right);
                     if (node->left)
                         next_layer.push(node->left);
                 } else {
                     if (node->left)
                         next_layer.push(node->left);
                     if (node->right)
                         next_layer.push(node->right);
                 }
             }
             ret.push_back(layer_val);
             swap(cur_layer, next_layer);
             dirt = !dirt;
         }
         return ret;
     }
 };