Leetcode 617 Merge Two Binary Trees 二叉树

题意:

     给定两棵树，将两棵树合并成一颗树

   输入

	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
输出 
合并的树
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

方法1不需要提供额外内存，但是时间居然很长
 1 /**
 2  * Definition for a binary tree node.
 3  * type TreeNode struct {
 4  *     Val int
 5  *     Left *TreeNode
 6  *     Right *TreeNode
 7  * }
 8  */
 9 func mergeTrees(t1 *TreeNode, t2 *TreeNode) *TreeNode {
10     if t1 == nil && t2 == nil{
11         return nil
12     }
13     if t1 != nil && t2 == nil{
14         return t1
15     }
16     if t1 == nil && t2 != nil{
17         return t2
18     }
19     if t1 != nil && t2 != nil{
20         t1.Val += t2.Val
21         t1.Left = mergeTrees(t1.Left, t2.Left);
22         t1.Right = mergeTrees(t1.Right, t2.Right);
23     }
24     return t1; 
25 }

 方法2需要提供额外内存，但是时间居然比方法1少，简直不可思议

 1 /**
 2  * Definition for a binary tree node.
 3  * type TreeNode struct {
 4  *     Val int
 5  *     Left *TreeNode
 6  *     Right *TreeNode
 7  * }
 8  */
 9 func newTreeNode() *TreeNode{
10     return &TreeNode{
11     }
12 }
13 
14 func dfs(m **TreeNode, t *TreeNode){
15     
16     if t == nil  {
17         return ;
18     }
19     
20     if (*m) == nil{
21         (*m) =  newTreeNode()
22     }
23     
24     (*m).Val += t.Val
25     dfs(&((*m).Left),  t.Left)
26     dfs(&((*m).Right), t.Right)
27   
28 }
29  
30 func mergeTrees(t1 *TreeNode, t2 *TreeNode) *TreeNode {
31     if t1 == nil && t2 == nil {
32         return nil
33     }
34     
35     m :=  newTreeNode()
36     
37     dfs(&m, t1);
38     dfs(&m, t2);
39     
40     return m; 
41 }

 

posted @ 2017-06-14 19:58  Breeze0806  阅读(1062)  评论(0)  编辑  收藏  举报