README.md

1379. 找出克隆二叉树中的相同节点

English Version

题目描述

给你两棵二叉树，原始树 original 和克隆树 cloned，以及一个位于原始树 original 中的目标节点 target。

其中，克隆树 cloned 是原始树 original 的一个 副本 。

请找出在树 cloned 中，与 target 相同 的节点，并返回对该节点的引用（在 C/C++ 等有指针的语言中返回 节点指针，其他语言返回节点本身）。

 

注意：你 不能 对两棵二叉树，以及 target 节点进行更改。只能 返回对克隆树 cloned 中已有的节点的引用。

 

示例 1:

输入: tree = [7,4,3,null,null,6,19], target = 3
输出: 3
解释: 上图画出了树 original 和 cloned。target 节点在树 original 中，用绿色标记。答案是树 cloned 中的黄颜色的节点（其他示例类似）。

示例 2:

输入: tree = [7], target =  7
输出: 7

示例 3:

输入: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4
输出: 4

 

提示：

• 树中节点的数量范围为 [1, 104] 。
• 同一棵树中，没有值相同的节点。
• target 节点是树 original 中的一个节点，并且不会是 null 。

 

进阶：如果树中允许出现值相同的节点，将如何解答？

解法

Python3

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


class Solution:
    def getTargetCopy(
        self, original: TreeNode, cloned: TreeNode, target: TreeNode
    ) -> TreeNode:
        res = None

        def dfs(original, cloned):
            nonlocal res
            if cloned is None:
                return
            if original == target:
                res = cloned
                return
            dfs(original.left, cloned.left)
            dfs(original.right, cloned.right)

        dfs(original, cloned)
        return res

Java

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

class Solution {
    private TreeNode res;

    public final TreeNode getTargetCopy(
        final TreeNode original, final TreeNode cloned, final TreeNode target) {
        dfs(original, cloned, target);
        return res;
    }

    private void dfs(TreeNode original, TreeNode cloned, TreeNode target) {
        if (cloned == null) {
            return;
        }
        if (original == target) {
            res = cloned;
            return;
        }
        dfs(original.left, cloned.left, target);
        dfs(original.right, cloned.right, target);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* res;

    TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) {
        dfs(original, cloned, target);
        return res;
    }

    void dfs(TreeNode* original, TreeNode* cloned, TreeNode* target) {
        if (!cloned) return;
        if (original == target) {
            res = cloned;
            return;
        }
        dfs(original->left, cloned->left, target);
        dfs(original->right, cloned->right, target);
    }
};

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 getTargetCopy(
    original: TreeNode | null,
    cloned: TreeNode | null,
    target: TreeNode | null,
): TreeNode | null {
    if (cloned === null) {
        return null;
    }
    if (cloned.val === target.val) {
        return cloned;
    }
    return (
        getTargetCopy(original, cloned.left, target) ||
        getTargetCopy(original, cloned.right, target)
    );
}

...