【LeetCode】617. 合并两个二叉树

最新推荐文章于 2025-01-12 11:51:27 发布

千筠Wyman

最新推荐文章于 2025-01-12 11:51:27 发布

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分类专栏： python 算法 LeetCode 文章标签： LeetCode 算法二叉树合并

本文链接：https://blog.youkuaiyun.com/sinat_36645384/article/details/103456063

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本文介绍了一种将两棵二叉树合并成一棵新二叉树的算法，当两棵树的节点重叠时，节点值相加；否则，取非空节点。提供了Python实现代码，演示了输入树和合并后的树的结构。

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问题描述

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.

假设有两棵二叉树，假设你用其中一棵树覆盖另一棵树，两棵树的一些节点是重叠的，而另一些不是。
你需要将它们合并到一个新的二叉树中。合并规则是，如果两个节点重叠，则将节点值相加作为合并节点的新值。否则，不为 null的节点将被用作新树的节点。

输入: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
输出: 
合并后的新树:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

Python 实现

# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution(object):
    def mergeTrees(self, t1, t2):
        """
        :type t1: TreeNode
        :type t2: TreeNode
        :rtype: TreeNode
        """
        if t1 == None:
            return t2
        elif t2 == None:
            return t1 
        else:
            tNode = TreeNode(t1.val + t2.val)
            tNode.left = self.mergeTrees(t1.left, t2.left)
            tNode.right = self.mergeTrees(t1.right, t2.right)
            return tNode

链接：https://leetcode.com/problems/merge-two-binary-trees/