【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;
    }
};