[leetcode] 103. Binary Tree Zigzag Level Order Traversal 解题报告

题目链接：https://leetcode.com/problems/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]
]

思路：可以用两个栈来分别保存奇数层和偶数层的结点，奇数层是从左到右，偶数层是从右到左。按照栈的后入先出的特性，在将结点加入到奇数层的栈的时候应该用右向左加，即先入栈根的右结点，再入栈根的左节点，这样计数层出栈的时候就会按照从左到右的顺序出栈。保存偶数层栈的顺序正好相反。

代码如下：

/**
 * 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) {
        if(!root) return result;
        st1.push(root);
        while(!st1.empty())
        {
            vector<int> tem;
            while(!st1.empty())//奇数层
            {
                TreeNode* node = st1.top();
                tem.push_back(node->val);
                st1.pop();
                if(node->left) st2.push(node->left);
                if(node->right) st2.push(node->right);
            }
            result.push_back(tem);
            tem.clear();
            while(!st2.empty())//偶数层
            {
                TreeNode* node = st2.top();
                tem.push_back(node->val);
                st2.pop();
                if(node->right) st1.push(node->right);
                if(node->left) st1.push(node->left);
            }
            if(tem.size() != 0) 
                result.push_back(tem);
        }
        return result;
    }
private:
    vector<vector<int>> result;
    stack<TreeNode*> st1;
    stack<TreeNode*> st2;
};