Leetcode[103]-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：使用层序遍历，借助队列，依次遍历每层，如果层数为偶数，则将该层的数字翻转，使用vector的reverse函数即可。

代码Code （C++）：

/**
 * 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> > res;
        if(!root) return res;
        vector<int> nums;
        queue<TreeNode *> que;
        TreeNode *p = root;
        que.push(p);

        int level  = 0; // curent level
        while(!que.empty()){
            nums.resize(0);
            level += 1;
            int queSize = que.size();
            for(int i = 0; i < queSize; i++ ) {
                p = que.front();
                nums.push_back(p->val);
                que.pop();
                if(p->left) que.push(p->left);
                if(p->right) que.push(p->right);
            }
            if(level % 2 == 0) reverse(nums.begin(),nums.end());
            res.push_back(nums);
        }
        return  res;         
    }
};