[LeetCode] 951. Flip Equivalent Binary Trees_Medium tag: DFS, divide and conquer

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 equivelent 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

Example 4:

Input: root1 = [0,null,1], root2 = []
Output: false

Example 5:

Input: root1 = [0,null,1], root2 = [0,1]
Output: true

 

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].

 

Ideas:

1. both None => true

2. both not None => root1.val == root2.val 并且recursive 来call children就好

3. 其他情况都是False

 

Code:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        if not root1 and not root2: return True
        elif root1 and root2 and root1.val == root2.val:
            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))
        else:
            return False