【Leetcode】Merge Two Binary Trees-优快云博客

本文链接：https://blog.youkuaiyun.com/wangzch7/article/details/82962958

本文详细解析了一道关于二叉树合并的经典算法题目。通过递归方法，当两棵树的节点重叠时，将节点值相加作为新节点值；否则，非空节点将用于构建新树。代码示例清晰展示了这一过程。

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题目：

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

解题思路：

这题目的意思是指给出两棵二叉树进行合并，可以采用递归的方法对左右子树分别合并即可

注意边界以及递归结束的条件

代码如下：

/**
 * 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;
        }
        if (t1 == NULL && t2 == NULL) {
            return NULL;
        }
        TreeNode* node = new TreeNode(t1->val + t2->val);
        node->left = mergeTrees(t1->left, t2->left);
        node->right = mergeTrees(t1->right, t2->right);
        return node;
    }
};