[Leetcode] 0951. Flip Equivalent Binary Trees

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內容目錄

Medium

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.

Example 1:

Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.

Example 2:

Input: root1 = [], root2 = []
Output: true

Example 3:

Input: root1 = [], root2 = [1]
Output: false

Constraints:

The number of nodes in each tree is in the range [0, 100].
Each tree will have unique node values in the range [0, 99].

Python

				
					# time complexity: O(n)
# space complexity: O(n)
from typing import Optional


class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


def traverse(node):
    if node is None:
        return
    if node.val:
        print(node.val)
    if node.left:
        traverse(node.left)
    if node.right:
        traverse(node.right)


class Solution:
    def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        if not root1 and not root2:
            return True
        if not root1 or not root2:
            return False
        if root1.val != root2.val:
            return False
        return (self.flipEquiv(root1.left, root2.left) and self.flipEquiv(root1.right, root2.right)) or (self.flipEquiv(root1.left, root2.right) and self.flipEquiv(root1.right, root2.left))


root1 = TreeNode(1)
root1.left = TreeNode(2)
root1.right = TreeNode(3)
root1.left.left = TreeNode(4)
root1.left.right = TreeNode(5)
root1.right.left = TreeNode(6)
root1.left.right.left = TreeNode(7)
root1.left.right.right = TreeNode(8)


root2 = TreeNode(1)
root2.left = TreeNode(3)
root2.right = TreeNode(2)
root2.left.right = TreeNode(6)
root2.right.left = TreeNode(4)
root2.right.right = TreeNode(5)
root2.right.right.left = TreeNode(8)
root2.right.right.right = TreeNode(7)
print(Solution().flipEquiv(root1, root2))

Medium

Python

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