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

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func mergeTrees(t1 *TreeNode, t2 *TreeNode) *TreeNode {
    if t1 == nil && t2 == nil{
        return nil
    }
    if t1 != nil && t2 == nil{
        return t1
    }
    if t1 == nil && t2 != nil{
        return t2
    }
    if t1 != nil && t2 != nil{
        t1.Val += t2.Val
        t1.Left = mergeTrees(t1.Left, t2.Left);
        t1.Right = mergeTrees(t1.Right, t2.Right);
    }
    return t1; 
}

 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

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func newTreeNode() *TreeNode{
    return &TreeNode{
    }
}

func dfs(m **TreeNode, t *TreeNode){
    
    if t == nil  {
        return ;
    }
    
    if (*m) == nil{
        (*m) =  newTreeNode()
    }
    
    (*m).Val += t.Val
    dfs(&((*m).Left),  t.Left)
    dfs(&((*m).Right), t.Right)
  
}
 
func mergeTrees(t1 *TreeNode, t2 *TreeNode) *TreeNode {
    if t1 == nil && t2 == nil {
        return nil
    }
    
    m :=  newTreeNode()
    
    dfs(&m, t1);
    dfs(&m, t2);
    
    return m; 
}