[LeetCode日记指南] 617. Merge Two Binary Trees（递归法）

最新推荐文章于 2024-12-08 10:56:24 发布

YorkChu

最新推荐文章于 2024-12-08 10:56:24 发布

阅读量184

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/chuyueh/article/details/88935520

leetcode 专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种合并两棵二叉树的方法。当两棵树的节点重叠时，将它们的值相加作为新树的值；否则，取非空节点作为新树的节点。采用递归方式从根节点开始处理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

思路

处理二叉树最简单的方法是递归，本题可以直接递归调用给定的函数，我们首先判断，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的结点值之和赋予t1，不用建立新的结点然后对t1和t2的左子结点调用递归并赋给t1的左子结点，再对t1和t2的右子结点调用递归并赋给t1的右子结点。递归调用全部出栈之后，返回的是t1的根结点。

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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;
    }
}