README.md

617. 合并二叉树

English Version

题目描述

给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。

你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。

示例 1:

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

注意: 合并必须从两个树的根节点开始。

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
        if root1 is None:
            return root2
        if root2 is None:
            return root1
        node = TreeNode(root1.val + root2.val)
        node.left = self.mergeTrees(root1.left, root2.left)
        node.right = self.mergeTrees(root1.right, root2.right)
        return node

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null) {
            return root2;
        }
        if (root2 == null) {
            return root1;
        }
        TreeNode node = new TreeNode(root1.val + root2.val);
        node.left = mergeTrees(root1.left, root2.left);
        node.right = mergeTrees(root1.right, root2.right);
        return node;
    }
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

 function mergeTrees(root1: TreeNode | null, root2: TreeNode | null): TreeNode | null {
    if (root1 == null && root2 == null) return null;
    if (root1 == null) return root2;
    if (root2 == null) return root1;
    let left = mergeTrees(root1.left, root2.left);
    let right = mergeTrees(root1.right, root2.right);
    return new TreeNode(root1.val + root2.val, left, right);
}; 

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        if (root1 == nullptr) {
            return root2;
        }
        if (root2 == nullptr) {
            return root1;
        }
        TreeNode *node = new TreeNode(root1->val + root2->val);
        node->left = mergeTrees(root1->left, root2->left);
        node->right = mergeTrees(root1->right, root2->right);
        return node;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {
	if root1 == nil {
		return root2
	}
	if root2 == nil {
		return root1
	}
	node := &TreeNode{Val: root1.Val + root2.Val}
	node.Left = mergeTrees(root1.Left, root2.Left)
	node.Right = mergeTrees(root1.Right, root2.Right)
	return node
}