알고리즘/LeetCode

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

앤테바 2022. 12. 23. 20:39
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문제)

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

 

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?

 

 

솔루션1)

- original을 순회할 때 cloned 트리도 함께 순회하면서 original에서 target node를 찾았다면 그 지점이 cloned에서도 찾은 노드에 해당됨

# 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 traverse(o_node, t_node):            
            if o_node is None:
                return None            
            
            if o_node.val == target.val:
                print('found')
                return t_node
            
            found_node = traverse(o_node.left, t_node.left)
            if found_node:
                return found_node
            
            found_node = traverse(o_node.right, t_node.right)
            if found_node:
                return found_node
            
            return None
        return traverse(original, cloned)

솔루션2)

- 솔루션1 코드 리팩토링

# 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 traverse(o_node, t_node):            
            if o_node is None:
                return None            
            
            if o_node == target:                
                return t_node
            
            l, r = traverse(o_node.left, t_node.left), traverse(o_node.right, t_node.right)
            return l if l else r
            
        return traverse(original, cloned)

솔루션3)

- 별도의 재귀함수를 만들지 않고 getTargetCopy() 재귀 함수로 사용해서 문제 해결

# 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:
        if original is None:
            return None
        
        if original == target:
            return cloned
        
        l = self.getTargetCopy(original.left, cloned.left, target)
        r = self.getTargetCopy(original.right, cloned.right, target)
        
        return l if l else r

 

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