【leetcode】617. Merge Two Binary Trees_leetcode617完整代码-优快云博客

本文链接：https://blog.youkuaiyun.com/xiaoquantouer/article/details/77184864

本文介绍了一种将两棵二叉树合并为一棵新二叉树的算法，当两棵树的节点重叠时，合并后的节点值等于两棵树对应节点值之和；否则，采用非空节点值。提供C++及Go语言实现。

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一、题目描述

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

Example 1:

Input: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
Output: 
Merged tree:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

Note: The merging process must start from the root nodes of both trees.

思路：用递归的思维去想，我的代码直接使用t1作为主树，值也是在上面修改了，最后返回t1，如果面试官说不能改动原来的树就新建一个node，然后赋值给这个node即可。

AC代码：

c++版

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if(t1 == NULL){
            return t2;
        }
        if(t2 == NULL){
            return t1;
        }
        t1->val += t2->val;
        t1->left = mergeTrees(t1->left, t2->left);
        t1->right = mergeTrees(t1->right, t2->right);
        return t1;
    }
};

golang版

/**
 * 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{
        return t2
    }
    if t2 == nil{
        return t1
    }
    t1.Val += t2.Val
    t1.Left = mergeTrees(t1.Left, t2.Left)
    t1.Right = mergeTrees(t1.Right, t2.Right)
    return t1

}