【leetcode】951. Flip Equivalent Binary Trees

发布时间：2020-12-14 04:28:25 所属栏目：大数据来源：网络整理

导读：题目如下： 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 som

题目如下：

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,7,8],root2 = [1,8,7] Output: true Explanation: We flipped at nodes with values 1,and 5. 
?

Note:

Each tree will have at most?100?nodes.

Each value in each tree will be a unique?integer in the range?[0,99].

解题思路：从两个根节点开始分别同步遍历两棵树，如果左右子树相同则表示不用交换；如果左右互相等于对方则交换；否则表示无法交换。

代码如下：

# 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
        else:
            return -1
    def traverse(self,node2):
        if node1 == None or node2 == None:
            return
        ret = self.verify(node1,node2)
        if ret == 0:
            self.traverse(node1.left,node2.left)
            self.traverse(node1.right,node2.right)
        elif ret == 1:
            node2.left,node2.right = node2.right,node2.left
            self.traverse(node1.left,node2.right)
        else:
            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
        self.traverse(root1,root2)
        return self.res

（编辑：李大同）

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