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;
 }