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1379. Find a Corresponding Node of a Binary Tree in a Clone of That Tree (M)

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Find a Corresponding Node of a Binary Tree in a Clone of That Tree (M)

题目

Given two binary trees original and cloned and given a reference to a node target in the original tree.

The cloned tree is a copy of the original tree.

Return a reference to the same node in the cloned tree.

Note that you are not allowed to change any of the two trees or the target node and the answer must be a reference to a node in the cloned tree.

Follow up: Solve the problem if repeated values on the tree are allowed.

Example 1:

技术图片

Input: tree = [7,4,3,null,null,6,19], target = 3
Output: 3
Explanation: In all examples the original and cloned trees are shown. The target node is a green node from the original tree. The answer is the yellow node from the cloned tree.

Example 2:

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Input: tree = [7], target =  7
Output: 7

Example 3:

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Input: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4
Output: 4

Example 4:

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Input: tree = [1,2,3,4,5,6,7,8,9,10], target = 5
Output: 5

Example 5:

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Input: tree = [1,2,null,3], target = 2
Output: 2 

Constraints:

  • The number of nodes in the tree is in the range [1, 10^4].
  • The values of the nodes of the tree are unique.
  • target node is a node from the original tree and is not null.

题意

给定两个树,其中一个是另一个的拷贝,要求在拷贝树中找到对应于原树中某一个结点的结点。

思路

DFS,对原树和拷贝树进行同步递归搜索即可。


代码实现

Java

class Solution {
    public final TreeNode getTargetCopy(final TreeNode original, final TreeNode cloned, final TreeNode target) {
        if (original == null || original == target) {
            return cloned;
        }

        TreeNode leftNode = getTargetCopy(original.left, cloned.left, target);
        if (leftNode != null) {
            return leftNode;
        }
        return getTargetCopy(original.right, cloned.right, target);
    }
}

1379. Find a Corresponding Node of a Binary Tree in a Clone of That Tree (M)

标签:mic   com   sam   nat   同步   origin   -o   amp   find   

原文地址:https://www.cnblogs.com/mapoos/p/14223547.html

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