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Formatted question description: https://leetcode.ca/all/1379.html

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

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:

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

Example 4:

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

Example 5:

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.

Related Topics:
Tree

Solution 1.

• Java
• C++
• Python
• TypeScript

• /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  
  class Solution {
      public final TreeNode getTargetCopy(final TreeNode original, final TreeNode cloned, final TreeNode target) {
          TreeNode copy = null;
          Queue<TreeNode> queue1 = new LinkedList<TreeNode>();
          Queue<TreeNode> queue2 = new LinkedList<TreeNode>();
          queue1.offer(original);
          queue2.offer(cloned);
          while (!queue1.isEmpty()) {
              TreeNode node1 = queue1.poll();
              TreeNode node2 = queue2.poll();
              if (node1 == target) {
                  copy = node2;
                  break;
              }
              TreeNode left1 = node1.left, right1 = node1.right;
              TreeNode left2 = node2.left, right2 = node2.right;
              if (left1 != null) {
                  queue1.offer(left1);
                  queue2.offer(left2);
              }
              if (right1 != null) {
                  queue1.offer(right1);
                  queue2.offer(right2);
              }
          }
          return copy;
      }
  }
  
  ############
  
  /**
   * 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);
      }
  }
  

• // OJ: https://leetcode.com/problems/find-a-corresponding-node-of-a-binary-tree-in-a-clone-of-that-tree/
  // Time: O(N)
  // Space: O(H)
  class Solution {
  public:
      TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) {
          if (!original) return nullptr;
          if (original == target) return cloned;
          auto ans = getTargetCopy(original->left, cloned->left, target);
          return ans ? ans : getTargetCopy(original->right, cloned->right, target);
      }
  };
  

• # 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
  
  
  

• /**
   * 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)
      );
  }
  
  

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