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  • 【leetcode】1457. Pseudo-Palindromic Paths in a Binary Tree

    题目如下:

    Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be pseudo-palindromic if at least one permutation of the node values in the path is a palindrome.

    Return the number of pseudo-palindromic paths going from the root node to leaf nodes. 

    Example 1:

    Input: root = [2,3,1,3,1,null,1]
    Output: 2 
    Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: 
    the red path [2,3,3], the green path [2,1,1], and the path [2,3,1]. Among these paths only red path and
    green path are pseudo-palindromic paths since the red path [2,3,3] can be rearranged in [3,2,3] (palindrome) and
    the green path [2,1,1] can be rearranged in [1,2,1] (palindrome).

    Example 2:

    Input: root = [2,1,1,1,3,null,null,null,null,null,1]
    Output: 1 
    Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: 
    the green path [2,1,1], the path [2,1,3,1], and the path [2,1]. Among these paths only the green path is pseudo-palindromic
    since [2,1,1] can be rearranged in [1,2,1] (palindrome).

    Example 3:

    Input: root = [9]
    Output: 1

    Constraints:

    • The given binary tree will have between 1 and 10^5 nodes.
    • Node values are digits from 1 to 9.

    解题思路:要构成伪回文串,需要满足一个条件,出现次数为奇数的数字最多只能有一个。所以,只需要在遍历树的过程中,记录遍历路径每个数字出现的次数即可。如果一个数字出现了偶数次,其实和出现0次是没有区别的。因此,我用二进制来保存每个数字出现的次数。

    代码如下:

    # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, val=0, left=None, right=None):
    #         self.val = val
    #         self.left = left
    #         self.right = right
    class Solution(object):
        res = 0
        def pseudoPalindromicPaths (self, root):
            """
            :type root: TreeNode
            :rtype: int
            """
            def recursive(node,num):
                num ^= pow(2,node.val - 1)
                if node.left == None and node.right == None:
                    bs = bin(num)[2:]
                    if bs.count('1') <= 1:
                        self.res += 1
                    return
                if node.left != None:
                    recursive(node.left,num)
                if node.right != None:
                    recursive(node.right,num)
    
            self.res = 0
            recursive(root,0)
            return self.res
            
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  • 原文地址:https://www.cnblogs.com/seyjs/p/13046942.html
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