首页 > 代码库 > 107. Binary Tree Level Order Traversal II

107. Binary Tree Level Order Traversal II

2024-10-06 23:29:38   213人阅读

题目：

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]
]
题目分析：
这道题本身不难，用一个广度优先遍历先分层，再将每层的结果自左向右输出。方便起见用到了栈和队列。记录这道题是一些语法的使用上，帮助自己回忆。
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  #include <stack>
11  #include <queue>
12 class Solution {
13 public:
14     struct Pair{
15         TreeNode* root;
16         int level;
17         Pair(TreeNode* _root,int _level) : root(_root),level(_level) {}
18     };
19     vector<vector<int>> levelOrderBottom(TreeNode* root) {
20         stack<Pair> s;
21         queue<Pair> q;
22         vector<vector<int>> v;
23         if(!root) return v;
24         q.push(Pair(root,0));
25         while(!q.empty())
26         {
27             if(q.front().root->right)
28                 q.push(Pair(q.front().root->right,q.front().level+1));
29             if(q.front().root->left)
30                 q.push(Pair(q.front().root->left,q.front().level+1));
31             s.push(q.front());
32             q.pop();
33         }
34         int level = s.top().level;
35         v=vector<vector<int>>(level+1);
36         while(!s.empty())
37         {
38             v[level-s.top().level].push_back(s.top().root->val);
39             s.pop();
40         }
41         return v;
42     }
43 };

 

107. Binary Tree Level Order Traversal II


联系
我们