binary-tree-zigzag-level-order-traversal——二叉树分层输出

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]
]

 

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

PS：二叉树分层输出，BFS并且记录翻转情况

 

/**
 * Definition for binary tree
 * 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==NULL) return res;
        queue<TreeNode*> q;
        q.push(root);
        bool reverse=false;
        while(!q.empty()){
            vector<int> v;
            int size=q.size();
            for(int i=0;i<size;++i){
                TreeNode *cur=q.front();
                q.pop();
                v.push_back(cur->val);
                if(cur->left!=NULL) q.push(cur->left);
                if(cur->right!=NULL) q.push(cur->right);
            }
            if(reverse){
                
                vector<int> tmp;
                for(int i=v.size()-1;i>=0;--i){
                    tmp.push_back(v[i]);
                }
                res.push_back(tmp);
            }else{
                res.push_back(v);
            }
            reverse=!reverse;
        }
        return res;
    }
};