Teaching Kids Programming – Find a Corresponding Node of a Binary Tree in a Clone of That Tree via Recursive Depth First Search Algorithm


Teaching Kids Programming: Videos on Data Structures and Algorithms

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.

find-the-node-in-the-cloned-binary-tree Teaching Kids Programming - Find a Corresponding Node of a Binary Tree in a Clone of That Tree via Recursive Depth First Search Algorithm algorithms DFS recursive teaching kids programming youtube video

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

Find a Corresponding Node of a Binary Tree in a Clone of That Tree via Recursive Depth First Search Algorithm

We can traverse the binary tree and its clone synchronously (at the same pace). We can Depth First Search, and the easiest way would be to use Recursion. The order of the traversal (Preorder, Inorder, Postorder) does not matter as long as all the nodes are visited.

With Recursive DFS, we can implement it as the following where the DFS function returns the node or None if not found.

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class Solution:
    def getTargetCopy(self, a: TreeNode, b: TreeNode, target: TreeNode) -> TreeNode:
        def dfs(a, b, t):
            if not a:
                return None
            if a.val == t.val:
                return b
            x = dfs(a.left, b.left, t)
            if x:
                return x
            return dfs(a.right, b.right, t)        
        return dfs(a, b, target)
class Solution:
    def getTargetCopy(self, a: TreeNode, b: TreeNode, target: TreeNode) -> TreeNode:
        def dfs(a, b, t):
            if not a:
                return None
            if a.val == t.val:
                return b
            x = dfs(a.left, b.left, t)
            if x:
                return x
            return dfs(a.right, b.right, t)        
        return dfs(a, b, target)

Also, we can set self.ans (Solution.ans) once we found the corresponding node.

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class Solution:
    def getTargetCopy(self, a: TreeNode, b: TreeNode, target: TreeNode) -> TreeNode:
        self.ans = None
        def dfs(a, b, t):
            if not a:
                return
            if a is t:
                self.ans = b
                return
            dfs(a.left, b.left, t)
            if self.ans:
                return
            dfs(a.right, b.right, t)        
        dfs(a, b, target)        
        return self.ans
class Solution:
    def getTargetCopy(self, a: TreeNode, b: TreeNode, target: TreeNode) -> TreeNode:
        self.ans = None
        def dfs(a, b, t):
            if not a:
                return
            if a is t:
                self.ans = b
                return
            dfs(a.left, b.left, t)
            if self.ans:
                return
            dfs(a.right, b.right, t)        
        dfs(a, b, target)        
        return self.ans

Both algorithms have O(N) time complexity and O(N) space due to Recursion. The Recursion will stop once the node is found.

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

–EOF (The Ultimate Computing & Technology Blog) —

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