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二叉树

617. 合并二叉树

English Version

题目描述

给你两棵二叉树： root1 和 root2 。

想象一下，当你将其中一棵覆盖到另一棵之上时，两棵树上的一些节点将会重叠（而另一些不会）。你需要将这两棵树合并成一棵新二叉树。合并的规则是：如果两个节点重叠，那么将这两个节点的值相加作为合并后节点的新值；否则，不为 null 的节点将直接作为新二叉树的节点。

返回合并后的二叉树。

注意: 合并过程必须从两个树的根节点开始。

 

示例 1：

输入：root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
输出：[3,4,5,5,4,null,7]

示例 2：

输入：root1 = [1], root2 = [1,2]
输出：[2,2]

 

提示：

  • 两棵树中的节点数目在范围 [0, 2000] 内
  • -104 <= Node.val <= 104

解法

方法一：递归

递归合并两棵树的节点。

  • 如果其中一棵树的当前节点为空，则返回另一棵树的当前节点作为合并后节点。
  • 如果两棵树的当前节点都不为空，则将它们的值相加作为合并后节点的新值，然后递归合并它们的左右子节点。

时间复杂度 $O(m)$，空间复杂度 $O(m)$。其中 $m$ 是两棵树的节点数的最小值。

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: Optional[TreeNode], root2: Optional[TreeNode]
    ) -> Optional[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;
    }
}

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) return root2;
        if (!root2) 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
}

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;
    const left = mergeTrees(root1.left, root2.left);
    const right = mergeTrees(root1.right, root2.right);
    return new TreeNode(root1.val + root2.val, left, right);
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn merge_trees(
        root1: Option<Rc<RefCell<TreeNode>>>,
        root2: Option<Rc<RefCell<TreeNode>>>,
    ) -> Option<Rc<RefCell<TreeNode>>> {
        match (root1.is_some(), root2.is_some()) {
            (false, false) => None,
            (true, false) => root1,
            (false, true) => root2,
            (true, true) => {
                {
                    let mut r1 = root1.as_ref().unwrap().borrow_mut();
                    let mut r2 = root2.as_ref().unwrap().borrow_mut();
                    r1.val += r2.val;
                    r1.left = Self::merge_trees(r1.left.take(), r2.left.take());
                    r1.right = Self::merge_trees(r1.right.take(), r2.right.take());
                }
                root1
            }
        }
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root1
 * @param {TreeNode} root2
 * @return {TreeNode}
 */
var mergeTrees = function (root1, root2) {
    if (!root1) {
        return root2;
    }
    if (!root2) {
        return root1;
    }
    const node = new TreeNode(root1.val + root2.val);
    node.left = mergeTrees(root1.left, root2.left);
    node.right = mergeTrees(root1.right, root2.right);
    return node;
};