107. Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

 

return its bottom-up level order traversal as:

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

将二叉树每一层的节点放入二维数组中

C++(6ms):
 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> levelOrderBottom(TreeNode* root) {
13         if (root == NULL)
14             return vector<vector<int>>() ;
15         vector<vector<int>> res ;
16         queue<TreeNode*> que ;
17         que.push(root) ;
18         while(!que.empty()){
19             int len = que.size() ;
20             vector<int> vec ;
21             for(int i = 0 ; i < len ; i++){
22                 TreeNode* node = que.front() ;
23                 que.pop() ;
24                 vec.push_back(node->val) ;
25                 if (node->left)
26                     que.push(node->left) ;
27                 if (node->right)
28                     que.push(node->right) ;
29             }
30             res.push_back(vec) ;
31         }
32         reverse(res.begin() , res.end()) ;
33         return res ;
34     }
35 };
 
原文地址:https://www.cnblogs.com/mengchunchen/p/8064069.html