Leetcode 107. Binary Tree Level Order Traversal II

107. Binary Tree Level Order Traversal II

Total Accepted: 85871 Total Submissions: 247487 Difficulty: Easy

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],

return its bottom-up level order traversal as:

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

思路：实质是求自底向上的层序情况。可以见该题的姐妹题题解：Leetcode 102. Binary Tree Level Order Traversal

代码：

 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<vector<int>> v;
14         if(root==NULL) return v;
15         queue<TreeNode*> q;
16         q.push(root);
17         TreeNode *begin,*end=root,*cur=root;
18         while(!q.empty()){
19             vector<int> temp;
20             begin=q.front();
21             q.pop();
22             while(begin!=end){
23                 temp.push_back(begin->val);
24                 if(begin->left){
25                     q.push(begin->left);
26                     cur=begin->left;
27                 }
28                 if(begin->right){
29                     q.push(begin->right);
30                     cur=begin->right;
31                 }
32                 begin=q.front();
33                 q.pop();
34             }
35             temp.push_back(begin->val);
36             if(begin->left){
37                 q.push(begin->left);
38                 cur=begin->left;
39             }
40             if(begin->right){
41                 q.push(begin->right);
42                 cur=begin->right;
43             }
44             v.insert(v.begin(),temp); //相较于102题，就是这里变动了
45             end=cur;
46         }
47         return v;
48     }
49 };