【dfs+位运算】1457. 二叉树中的伪回文路径

https://leetcode.cn/problems/pseudo-palindromic-paths-in-a-binary-tree/

/**
 * 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:
    int res = 0;
    bool isPseudoPa(unordered_map<int,int> &curPath) {
        int odd = 0;
        for (auto it = curPath.begin(); it != curPath.end(); it++) {
            if (it->second %2) {
                odd++;
            }
        }
        if (odd <= 1) return true;
        return false;
    }
    void dfs(TreeNode* cur, unordered_map<int,int> &curPath) {
        if (cur == nullptr) {
            return;
        }
        curPath[cur->val]++; 
        if (cur->left == nullptr && cur->right == nullptr) {
            res += isPseudoPa(curPath);
            curPath[cur->val]--; 
            return;
        }
        dfs(cur->left, curPath);
        dfs(cur->right, curPath);
        curPath[cur->val]--; 

    }
    int pseudoPalindromicPaths (TreeNode* root) {
        unordered_map<int,int> mp;
        dfs(root, mp);
        return res;
    }
};

/**
 * 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:
    int res = 0;
    void dfs(TreeNode* cur, array<int, 10> &curPath) {
        if (cur == nullptr) {
            return;
        }
        curPath[cur->val] ^= 1; 
        if (cur->left == nullptr && cur->right == nullptr) {
            res += accumulate(curPath.begin(), curPath.end(), 0) <= 1;
            curPath[cur->val] ^= 1; 
            return;
        }

        dfs(cur->left, curPath);
        dfs(cur->right, curPath);
        curPath[cur->val] ^= 1; 

    }
    int pseudoPalindromicPaths (TreeNode* root) {
        array<int, 10> p{};
        dfs(root, p);
        return res;
    }
};

/**
 * 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:
    int res = 0;
    void dfs(TreeNode* cur, int mask) {
        if (cur == nullptr) {
            return;
        }
        mask ^= 1 << cur->val; 
        if (cur->left == nullptr && cur->right == nullptr) {
            res += (mask == (mask & -mask) ? 1 : 0);
            return;
        }
        dfs(cur->left, mask);
        dfs(cur->right, mask);
    }
    int pseudoPalindromicPaths (TreeNode* root) {

        dfs(root, 0);
        return res;
    }
};