[LeetCode] 617. Merge Two Binary Tree-优快云博客

本文介绍了一种将两棵可能部分重叠的二叉树合并为一棵新二叉树的算法。合并规则为：若有节点重叠，则将其值相加作为新节点值；若不重叠，则使用非空节点值。通过递归方式从根节点开始合并。

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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

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

思路：

首先判断，如果t1不存在，则直接返回t2，反之，如果t2不存在，则直接返回t1。如果上面两种情况都不满足，那么以t1和t2的结点值之和建立新结点t，然后对t1和t2的左子结点调用递归并赋给t的左子结点，再对t1和t2的右子结点调用递归并赋给t的右子结点，返回t结点即可

代码：

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if (!t1) return t2;
        if (!t2) return t1;
        TreeNode *t = new TreeNode(t1->val + t2->val);
        t->left = mergeTrees(t1->left, t2->left);
        t->right = mergeTrees(t1->right, t2->right);
        return t;
    }
};