【LeetCode 113】 Path Sum II

題目描述

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

代碼

/**
 * 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:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res;
        vector<int> curs;
        dfs(root, curs, sum, 0, res);
        return res;
    }
    
    void dfs(TreeNode* root, vector<int>& curs, int sum, int cur, vector<vector<int>>& res) {
        if (root == NULL) return;
        if (root->left == NULL && root->right == NULL && cur+root->val == sum) {
            curs.push_back(root->val);
            vector<int> tmp(curs.begin(), curs.end());
            curs.pop_back();
            res.push_back(tmp);
            return;
        }
        
        if (root->left) {
            curs.push_back(root->val);
            dfs(root->left, curs, sum, cur+root->val, res);
            curs.pop_back();
        }
        if (root->right){
            curs.push_back(root->val);
            dfs(root->right, curs, sum, cur+root->val, res);
            curs.pop_back();
        }
        return;
    }
};