103-Binary Tree Zigzag Level Order Traversal

题目

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,null,null,15,7]

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

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

分析

访问顺序分为奇数行和偶数行。

实现

class Solution {public:vector<vector<int>> zigzagLevelOrder(TreeNode * root)    {        vector<vector<int>> res;        if (!root) return res;        vector<int> tmpRes;        TreeNode *curr = NULL;        stack<TreeNode* > stack1, stack2;        stack1.push(root);        // stack1 保存奇数层的结点         // stack2 保存偶数层的结点        while (!stack1.empty() || !stack2.empty())        {            //访问奇数层的结点，并将下一层的结点存入stack2            while (!stack1.empty())            {                curr = stack1.top();                tmpRes.push_back(curr->val);                stack1.pop();                if (curr->left)                    stack2.push(curr->left);                if (curr->right)                    stack2.push(curr->right);            }            if(!tmpRes.empty())                res.push_back(tmpRes);            tmpRes.clear();            //访问偶数层的结点，并将下一层的结点存入stack1            while (!stack2.empty())            {                curr = stack2.top();                tmpRes.push_back(curr->val);                stack2.pop();                if (curr->right)                    stack1.push(curr->right);                if (curr->left)                    stack1.push(curr->left);            }            if(!tmpRes.empty())                res.push_back(tmpRes);            tmpRes.clear();        }        return res;    }};