【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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

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

二叉数之字形输出。跟上一题很像102. Binary Tree Level Order Traversal，上一题的层序遍历，每一层都是从左到右输出，这道题每一层的输出变成了从左到右或从右到左。每层输出时判断一下标志，倒置一下就好了。

/**
 * 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;
        queue<TreeNode*> que;
        if(root==NULL) return res;
        bool flag=true;
        que.push(root);
        
        while(!que.empty()){
            vector<int> onelevel;
            int size = que.size();
            for(int i=0;i<size;i++){
                TreeNode* tre = que.front();
                que.pop();
                onelevel.push_back(tre->val);
                if(tre->left) que.push(tre->left);
                if(tre->right) que.push(tre->right);
            }
            //这里多了一个判断，是不是倒序输出
            if(flag){
                flag=!flag;
                res.push_back(onelevel);
            }
            else{
                flag=!flag;
                reverse(onelevel.begin(),onelevel.end());
                res.push_back(onelevel);
            }
        }
        return res;
    }
};