[LeetCode]Binary Tree Level Order Traversal

Binary Tree Level Order Traversal

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

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

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

 

BFS，广度优先搜索。递归实现。迭代版需要用到队列。

 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     void bfs(TreeNode* root,int level,vector<vector<int>>& result)
13     {
14         if(!root) return;
15         if(level>result.size()) result.push_back(vector<int>());
16         result[level-1].push_back(root->val);
17         bfs(root->left,level+1,result);
18         bfs(root->right,level+1,result);
19     }
20     vector<vector<int>> levelOrder(TreeNode* root) {
21         vector<vector<int>> result;
22         bfs(root,1,result);
23         return result;
24     }
25 };

 

posted @ 2015-09-14 16:42  Sean_le  阅读(120)  评论(0)  编辑  收藏  举报