Leetcode - path sum系列

path sum I

DFS+判断
https://leetcode.com/problems/path-sum/

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (!root) return false;
        return dfs(root, sum);
    }
    
    bool dfs(TreeNode *root, int sum) {
        if (!root) return false;
        sum -= root->val;
        if (!root->right && !root->left) return sum == 0;
        return dfs(root->right, sum) | dfs(root->left, sum);
    }
};

path sum II

dfs + backtracking, 如何恢复现场

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res; vector<int> v;
        dfs(root, sum, v, res);
        return res;
    }
    
    void dfs(TreeNode *root, int sum, vector<int>& v, vector<vector<int>>& res) {
        if (!root) return;
        sum -= root->val;
        v.push_back(root->val);
        if (sum == 0 && !root->left && !root->right) res.push_back(v);
        dfs(root->left, sum, v, res);
        dfs(root->right, sum, v, res);
        v.pop_back();
    }
};

path sum III

每个点都要dfs俩次,一次是remaining sum, 一次是sum。 因此help要递归,主函数也要递归。 => 用queue操作也ok

/**
 * 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 {
    int cnt = 0;
public:
    int pathSum(TreeNode* root, int sum) {
        if (!root) return 0;

        dfs(root, sum, cnt);
        pathSum(root->left, sum);
        pathSum(root->right, sum);
        return cnt;
    }
    
    void dfs(TreeNode *root, int sum, int & cnt) {
        if (!root) return;
        if (root->val == sum) cnt++;
        dfs(root->left, sum-root->val, cnt);
        dfs(root->right, sum-root->val, cnt);
    }
};
posted @ 2020-10-16 04:38  linsinan1995  阅读(49)  评论(0)    收藏  举报