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

Description

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.

Example 1:

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:

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

Example 3:

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

Constraints:

The number of nodes in the tree is in the range [1, 10⁴].
The values of the nodes of the tree are unique.
target node is a node from the original tree and is not null.

Follow up: Could you solve the problem if repeated values on the tree are allowed?

Solutions

Solution 1: DFS

We design a function $dfs(root1, root2)$, which performs DFS traversal simultaneously in trees $root1$ and $root2$. When traversing to a node, if this node happens to be $target$, then we return the corresponding node in $root2$. Otherwise, we recursively search for $target$ in the left and right subtrees of $root1$ and $root2$, and return the result that is not empty.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the tree.

Python3JavaC++TypeScriptC#

# 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:
        def dfs(root1: TreeNode, root2: TreeNode) -> TreeNode:
            if root1 is None:
                return None
            if root1 == target:
                return root2
            return dfs(root1.left, root2.left) or dfs(root1.right, root2.right)

        return dfs(original, cloned)

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

class Solution {
    private TreeNode target;

    public final TreeNode getTargetCopy(
        final TreeNode original, final TreeNode cloned, final TreeNode target) {
        this.target = target;
        return dfs(original, cloned);
    }

    private TreeNode dfs(TreeNode root1, TreeNode root2) {
        if (root1 == null) {
            return null;
        }
        if (root1 == target) {
            return root2;
        }
        TreeNode res = dfs(root1.left, root2.left);
        return res == null ? dfs(root1.right, root2.right) : res;
    }
}

/**
 * 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* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) {
        function<TreeNode*(TreeNode*, TreeNode*)> dfs = [&](TreeNode* root1, TreeNode* root2) -> TreeNode* {
            if (root1 == nullptr) {
                return nullptr;
            }
            if (root1 == target) {
                return root2;
            }
            TreeNode* left = dfs(root1->left, root2->left);
            return left == nullptr ? dfs(root1->right, root2->right) : left;
        };
        return dfs(original, cloned);
    }
};

/**
 * 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 {
    const dfs = (root1: TreeNode | null, root2: TreeNode | null): TreeNode | null => {
        if (!root1) {
            return null;
        }
        if (root1 === target) {
            return root2;
        }
        return dfs(root1.left, root2.left) || dfs(root1.right, root2.right);
    };
    return dfs(original, cloned);
}

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

public class Solution {
    private TreeNode target;

    public TreeNode GetTargetCopy(TreeNode original, TreeNode cloned, TreeNode target) {
        this.target = target;
        return dfs(original, cloned);
    }

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

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

Description

Solutions

Solution 1: DFS

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