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1485. Clone Binary Tree With Random Pointer πŸ”’

DifficultyMedium

Description

A binary tree is given such that each node contains an additional random pointer which could point to any node in the tree or null.

Return a deep copy of the tree.

The tree is represented in the same input/output way as normal binary trees where each node is represented as a pair of [val, random_index] where:

  • val: an integer representing Node.val
  • random_index: the index of the node (in the input) where the random pointer points to, or null if it does not point to any node.

You will be given the tree in class Node and you should return the cloned tree in class NodeCopy. NodeCopy class is just a clone of Node class with the same attributes and constructors.

 

Example 1:

Input: root = [[1,null],null,[4,3],[7,0]]
Output: [[1,null],null,[4,3],[7,0]]
Explanation: The original binary tree is [1,null,4,7].
The random pointer of node one is null, so it is represented as [1, null].
The random pointer of node 4 is node 7, so it is represented as [4, 3] where 3 is the index of node 7 in the array representing the tree.
The random pointer of node 7 is node 1, so it is represented as [7, 0] where 0 is the index of node 1 in the array representing the tree.

Example 2:

Input: root = [[1,4],null,[1,0],null,[1,5],[1,5]]
Output: [[1,4],null,[1,0],null,[1,5],[1,5]]
Explanation: The random pointer of a node can be the node itself.

Example 3:

Input: root = [[1,6],[2,5],[3,4],[4,3],[5,2],[6,1],[7,0]]
Output: [[1,6],[2,5],[3,4],[4,3],[5,2],[6,1],[7,0]]

 

Constraints:

  • The number of nodes in the tree is in the range [0, 1000].
  • 1 <= Node.val <= 106

Solutions

Solution 1: Hash Table + DFS

Thinking

Besides left and right there is a \(\textit{random}\) pointer that may form cycles, so a naive recursion can loop or copy a node twice.

Map each original node to its copy: create the copy first, then recurse on left, right, and \(\textit{random}\). A hit in the map returns the existing copy.

We use a hash table \(\textit{seen}\) to record the correspondence between each node in the original tree and its copy, then perform a depth-first search.

Define \(\text{dfs}(root)\) to return the copy of \(root\):

  • If \(root\) is null, return null;
  • If \(root\) is already in \(\textit{seen}\), return \(\textit{seen}[root]\);
  • Otherwise create a copy node \(\textit{copy}\), set \(\textit{seen}[root] = \textit{copy}\), then recurse on the left child, right child, and \(\textit{random}\) pointer;
  • Finally return \(\textit{copy}\).

The main function returns \(\text{dfs}(root)\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the number of nodes.

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# Definition for Node.
# class Node:
#     def __init__(self, val=0, left=None, right=None, random=None):
#         self.val = val
#         self.left = left
#         self.right = right
#         self.random = random


class Solution:
    def copyRandomBinaryTree(self, root: "Optional[Node]") -> "Optional[NodeCopy]":
        def dfs(root: Optional[Node]) -> Optional[NodeCopy]:
            if root is None:
                return None
            if root in seen:
                return seen[root]
            copy = NodeCopy(root.val)
            seen[root] = copy
            copy.left = dfs(root.left)
            copy.right = dfs(root.right)
            copy.random = dfs(root.random)
            return copy

        seen = {}
        return dfs(root)

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/**
 * Definition for Node.
 * public class Node {
 *     int val;
 *     Node left;
 *     Node right;
 *     Node random;
 *     Node() {}
 *     Node(int val) { this.val = val; }
 *     Node(int val, Node left, Node right, Node random) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *         this.random = random;
 *     }
 * }
 */

class Solution {
    private Map<Node, NodeCopy> seen;

    public NodeCopy copyRandomBinaryTree(Node root) {
        seen = new HashMap<>();
        return dfs(root);
    }

    private NodeCopy dfs(Node root) {
        if (root == null) {
            return null;
        }
        if (seen.containsKey(root)) {
            return seen.get(root);
        }
        NodeCopy copy = new NodeCopy(root.val);
        seen.put(root, copy);
        copy.left = dfs(root.left);
        copy.right = dfs(root.right);
        copy.random = dfs(root.random);
        return copy;
    }
}

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/**
 * Definition for a Node.
 * struct Node {
 *     int val;
 *     Node *left;
 *     Node *right;
 *     Node *random;
 *     Node() : val(0), left(nullptr), right(nullptr), random(nullptr) {}
 *     Node(int x) : val(x), left(nullptr), right(nullptr), random(nullptr) {}
 *     Node(int x, Node *left, Node *right, Node *random) : val(x), left(left), right(right), random(random) {}
 * };
 */

class Solution {
public:
    NodeCopy* copyRandomBinaryTree(Node* root) {
        unordered_map<Node*, NodeCopy*> seen;
        auto dfs = [&](this auto&& dfs, Node* root) -> NodeCopy* {
            if (!root) {
                return nullptr;
            }
            if (seen.contains(root)) {
                return seen[root];
            }
            NodeCopy* copy = new NodeCopy(root->val);
            seen[root] = copy;
            copy->left = dfs(root->left);
            copy->right = dfs(root->right);
            copy->random = dfs(root->random);
            return copy;
        };
        return dfs(root);
    }
};

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/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Left *Node
 *     Right *Node
 *     Random *Node
 * }
 */

func copyRandomBinaryTree(root *Node) *NodeCopy {
    seen := make(map[*Node]*NodeCopy)
    var dfs func(root *Node) *NodeCopy
    dfs = func(root *Node) *NodeCopy {
        if root == nil {
            return nil
        }
        if v, ok := seen[root]; ok {
            return v
        }
        copy := &NodeCopy{Val: root.Val}
        seen[root] = copy
        copy.Left = dfs(root.Left)
        copy.Right = dfs(root.Right)
        copy.Random = dfs(root.Random)
        return copy
    }
    return dfs(root)
}

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/**
 * Definition for Node.
 * class Node {
 *     val: number
 *     left: Node | null
 *     right: Node | null
 *     random: Node | null
 *     constructor(val?: number, left?: Node | null, right?: Node | null, random?: Node | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *         this.random = (random===undefined ? null : random)
 *     }
 * }
 */

function copyRandomBinaryTree(root: Node | null): NodeCopy | null {
    const seen = new Map<Node, NodeCopy>();
    const dfs = (root: Node | null): NodeCopy | null => {
        if (root === null) {
            return null;
        }
        if (seen.has(root)) {
            return seen.get(root)!;
        }
        const copy = new NodeCopy(root.val);
        seen.set(root, copy);
        copy.left = dfs(root.left);
        copy.right = dfs(root.right);
        copy.random = dfs(root.random);
        return copy;
    };
    return dfs(root);
}

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/*
// Definition for a Node.
public class Node {
    public int val;
    public Node left;
    public Node right;
    public Node random;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _random) {
        val = _val;
        left = _left;
        right = _right;
        random = _random;
    }
}
*/

public class Solution {
    public NodeCopy CopyRandomBinaryTree(Node root) {
        var seen = new Dictionary<Node, NodeCopy>();
        NodeCopy Dfs(Node root) {
            if (root == null) {
                return null;
            }
            if (seen.ContainsKey(root)) {
                return seen[root];
            }
            var copy = new NodeCopy(root.val);
            seen[root] = copy;
            copy.left = Dfs(root.left);
            copy.right = Dfs(root.right);
            copy.random = Dfs(root.random);
            return copy;
        }
        return Dfs(root);
    }
}

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