【LeetCode】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]
]

链接：https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/

题解：对树进行bfs即可，这里要每层加入顺序交替变换，正常思路就是遍历时也要交替变换，实际上这里结果要求的只是把每层的值存进来即可，所以我们遍历依旧顺序进行，然后加入数组的index进行变换即可，做到这一点只需要每个队列的长度记录下来，还有空指针用nullptr，刚开始时候加入左右孩子没有判断是否为空一直报错。。下面100%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) {
        if (root == nullptr) {
            return vector<vector<int> > ();
        }
        vector<vector<int>> ans;
        queue<TreeNode*> q;
        q.push(root);
        bool flag = true;
        
        while(!q.empty()){
            int len = q.size();
            vector<int> tmp(len);
            for(int i = 0;i < len;i++){
                TreeNode* p = q.front();
                q.pop();
                int index = (flag)? i : (len-i-1);
                tmp[index] = p->val;
                if(p->left)
                    q.push(p->left);
                if(p->right)
                    q.push(p->right);
            }
            
            flag = !flag;
            ans.push_back(tmp);
            
        }
        
        return ans;
    }
};