1457. Pseudo-Palindromic Paths in a Binary Tree

叁歲伎倆 2024-04-08 09:24 70阅读 0赞

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:**

![20fea6785473432daca6a8742de9de3d.png][]

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:**

![d877326b3e9f74dad6102f17d490cdb3.png][]

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

题目：给定一个二叉树，每个节点都是由1~9的数字组成，问能组成伪回文的线路有几条。

思路：由于节点是由1~9的数字组成，用一个长度为10的vector保存每条路个数字出现的次数。如果出现次数为奇数次的数字数量小于等于 1，则证明该条路可以组成伪回文，结果加1. 代码：

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
     *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
     *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
     * };
     */
    class Solution {
    public:
        void helper(TreeNode* node, vector<int>& record, int& res){
            if(!node) return;
            record[node->val]++;
            if(!node->left && !node->right){
                int odd = 0;
                for(int i = 1; i < 10; i++){
                    if(record[i] % 2 == 1) odd++;
                }
                if(odd <= 1) res++;
            }
            if(node->left) helper(node->left, record, res);
            if(node->right) helper(node->right,record, res);
            record[node->val]--;
        }
        int pseudoPalindromicPaths (TreeNode* root) {
            int res = 0;
            vector<int> record(10, 0);
            helper(root, record, res);
            return res;
        }
    };

[20fea6785473432daca6a8742de9de3d.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/08/44bcd3c2415d4eab82c5040301d440fe.png
[d877326b3e9f74dad6102f17d490cdb3.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/04/08/ed7eddd25bb94768b1087eb051d4adb9.png