Flip equivalent binary trees [DFS]¶
Time: O(N); Space: O(H); 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.
Write a function that determines whether two binary trees are flip equivalent. The trees are given by root nodes root1 and root2.
Example 1:
Input: root1 = {TreeNode} [1,2,3,4,5,6,null,null,null,7,8], root2 = {TreeNode} [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.
Flipped Trees Diagram
Notes:
Each tree will have at most 100 nodes.
Each value in each tree will be a unique integer in the range [0, 99].
[1]:
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
1. Depth First Search¶
[2]:
class Solution1(object):
"""
Time: O(N)
Space: O(H)
"""
def flipEquiv(self, root1, root2):
"""
:type root1: TreeNode
:type root2: TreeNode
:rtype: bool
"""
if not root1 and not root2:
return True
if not root1 or not root2 or 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))
[3]:
s = Solution1()
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.right.right = TreeNode()
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.left = TreeNode()
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)
assert s.flipEquiv(root1, root2) == True