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Leetcode 107. Binary Tree Level Order Traversal II

时间:2017-02-18 14:29:06      阅读:138      评论:0      收藏:0      [点我收藏+]

标签:begin   struct   log   des   str   mil   null   leetcode   blog   

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]
]

 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         vector<int> row;
14         vector<vector<int>> v;
15         if(root == NULL)
16             return v;
17         queue<TreeNode*> qu;
18         TreeNode* node;
19         qu.push(root);
20         while(!qu.empty()){
21             int l = qu.size();
22             for(int i = 0; i < l; i++){
23                  node = qu.front();
24                  qu.pop();
25                  row.push_back(node -> val);
26                  if(node -> left != NULL)
27                     qu.push(node -> left);
28                 if(node -> right != NULL)
29                     qu.push(node -> right);
30             }
31             v.insert(v.begin(), row);
32             row.clear();
33         }
34         return v;
35     }
36 };

 

 

Leetcode 107. Binary Tree Level Order Traversal II

标签:begin   struct   log   des   str   mil   null   leetcode   blog   

原文地址:http://www.cnblogs.com/qinduanyinghua/p/6413025.html

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