【leetcode】951. Flip Equivalent Binary Trees(示例代码)

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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 = [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.



  1. Each tree will have at most 100 nodes.
  2. Each value in each tree will be a unique integer in the range [0, 99].



# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    res = True
    def verify(self,node1,node2):
        leftV1 = None if node1.left == None else node1.left.val
        leftV2 = None if node2.left == None else node2.left.val
        rightV1 = None if node1.right == None else node1.right.val
        rightV2 = None if node2.right == None else node2.right.val
        if leftV1 == leftV2 and rightV1 == rightV2:
            return 0
        elif leftV1 == rightV2 and rightV1 == leftV2:
            return 1
            return -1
    def traverse(self,node1,node2):
        if node1 == None or node2 == None:
        ret = self.verify(node1,node2)
        if ret == 0:
            self.traverse(node1.right, node2.right)
        elif ret == 1:
            node2.left,node2.right = node2.right,node2.left
            self.traverse(node1.left, node2.left)
            self.traverse(node1.right, node2.right)
            self.res = False

    def flipEquiv(self, root1, root2):
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        if (root1 == None) ^ (root2 == None):
            return False
        self.res = True
        return self.res



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