Teaching Kids Programming – Breadth First Search Algorithm to Find a Corresponding Node of a Binary Tree in a Clone of That Tree

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.

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 both trees at the same step with Breadth First Search Algorithm (BFS). The BFS visits each node in the tree (or graph) in the order of levels. It requires a queue which we can use deque in Python to achieve O(1) constant in both enque and deque operations.

We push the node of the tree and its clone as a tuple in the queue. We can deque one node at a time or we can deque all nodes in the queue to ensure all the nodes in the queue at anytime are on the same level.

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class Solution:
    def getTargetCopy(self, a: TreeNode, b: TreeNode, target: TreeNode) -> TreeNode:
        q = deque([(a, b)])
        while q:
            #n = len(q)
            #for _ in range(n):
            x, y = q.popleft()
            if x is target:
                return y
            if x.left:
                q.append((x.left, y.left))
            if y.right:
                q.append((x.right, y.right))

class Solution:
    def getTargetCopy(self, a: TreeNode, b: TreeNode, target: TreeNode) -> TreeNode:
        q = deque([(a, b)])
        while q:
            #n = len(q)
            #for _ in range(n):
            x, y = q.popleft()
            if x is target:
                return y
            if x.left:
                q.append((x.left, y.left))
            if y.right:
                q.append((x.right, y.right))

The time complexity is O(N) as we need to visit each node exactly at most once. And the space complexity is O(N) as we need a queue.

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

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

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