dfs

leetcode 102424

python/951.flip-equivalent-binary-trees.py

@@ -0,0 +1,99 @@
+#
+# @lc app=leetcode id=951 lang=python3
+#
+# [951] Flip Equivalent Binary Trees
+#
+# https://leetcode.com/problems/flip-equivalent-binary-trees/description/
+#
+# algorithms
+# Medium (67.12%)
+# Likes:    2328
+# Dislikes: 96
+# Total Accepted:    156.1K
+# Total Submissions: 232.1K
+# Testcase Example:  '[1,2,3,4,5,6,null,null,null,7,8]\n[1,3,2,null,6,4,5,null,null,null,null,8,7]'
+#
+# 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].
+#
+#
+#
+
+
+# @lc code=start
+# 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
+# Definition for a binary tree node.
+from typing import Optional
+
+
+class Solution:
+    def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
+
+        return self.check(root1, root2)
+
+    def check(self, node1, node2):
+
+        if not node1 and node2:
+            return False
+
+        if node1 and not node2:
+            return False
+
+        if not node1 and not node2:
+            return True
+
+        if node1.val == node2.val:
+            return (
+                self.check(node1.left, node2.left)
+                and self.check(node1.right, node2.right)
+            ) or (
+                self.check(node1.left, node2.right)
+                and self.check(node1.right, node2.left)
+            )
+
+        return False
+
+
+# @lc code=end