[LeetCode102] Binary Tree Level Order Traversal

题目：

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

 

return its level order traversal as:

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

分类：Tree BFS 

代码：

递归：

 1 class Solution {
 2 private:
 3     //递归
 4     vector<vector<int>> res;
 5     void dfs(TreeNode* root, int height)
 6     {
 7         if(!root)
 8             return;
 9         while(res.size() <= height)
10             res.push_back(vector<int>());
11         res[height].push_back(root->val);
12         dfs(root->left,height+1);
13         dfs(root->right,height+1);
14     }
15 public:
16     vector<vector<int>> levelOrder(TreeNode* root) {      
17         //递归
18         dfs(root,0);
19         return res;
20     }
21 };

非递归：

 1 vector<vector<int>> levelOrder(TreeNode* root) {
 2         vector<vector<int>> ans;
 3         vector<int> temp;
 4         if(!root)
 5             return ans;
 6         queue<TreeNode*> que;
 7         que.push(root);
 8         int parentSize = 1, childSize = 0;
 9         while(!que.empty())
10         {
11             auto node = que.front();
12             que.pop();
13             temp.push_back(node->val);
14             if(node->left)
15             {
16                 ++childSize;
17                 que.push(node->left);
18             }
19             if(node->right)
20             {
21                 ++childSize;
22                 que.push(node->right);
23             }
24             --parentSize;
25             if(parentSize == 0)
26             {
27                 ans.push_back(temp);
28                 temp.clear();
29                 parentSize = childSize;
30                 childSize = 0;
31             }
32         }
33         return ans;
34     }

 

posted @ 2016-08-16 10:55  zhangbaochong  阅读(180)  评论(0)  编辑  收藏  举报