113-Path Sum II

类别：DFS
难度：medium

题目描述

算法分析

与112的算法基本一样，但是要注意在实现的过程中对于左子树和右子树path分别进行复制，因为if是先后进行的，第一个if判断后会在第二个if中使用path，而此时path已经发生改变

代码实现

class Solution {public:    void findPath(vector<vector<int>>& result, vector<int> path, TreeNode* root, int& sum) {        if (root->left == NULL && root->right == NULL) {            int n = path.size();            int count = 0;            for (int i = 0; i < n; ++i) {                count += path[i];            }            if (count == sum) {                result.push_back(path);            }        }        //  notice here copy the path, because after judge root->left, it still use path to judge root->right        vector<int> leftPath = path;        vector<int> rightPath = path;        if (root->left != NULL) {            leftPath.push_back(root->left->val);            findPath(result, leftPath, root->left, sum);        }        if (root->right != NULL) {            rightPath.push_back(root->right->val);            findPath(result, rightPath, root->right, sum);        }    }    vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int>> result;        if (root == NULL) return result;        vector<int> path;        path.push_back(root->val);        findPath(result, path, root, sum);    }};