113. Path Sum II

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

For 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]]

【问题分析】

  这个问题和Path Sum类似，提高了一些要求：

  1、结束条件不同，找到满足条件的路径后不能直接结束，需要继续寻找可能满足条件的路径
  2、要求把找到的路径保存下来

【AC代码】

  

class Solution {public:    vector<vector<int>> pathSum(TreeNode* root, int sum) {            std::vector<int> path;            std::vector<std::vector<int> > paths;            if (root == NULL) {                      //返回所有结果路径集合                //paths.push_back(path);               //the key point               return paths;             }            if (root->left == NULL && root->right == NULL) {    //遇到叶子节点，获取当前路径，并加入结果路径集合中                if (root->val == sum) {                    path.push_back(sum);                    paths.push_back(path);                }                return paths;            }            if (root->left != NULL) {                std::vector<std::vector<int> > lpaths = pathSum(root->left, sum - root->val);    //获取左子节点的所有路径                for (int i = 0; i < lpaths.size(); ++i) {                                        //将当前节点插入到每条路径的起始位置                    lpaths[i].insert(lpaths[i].begin(), root->val);                    paths.push_back(lpaths[i]);                }            }            if (root->right != NULL) {                std::vector<std::vector<int> > rpaths = pathSum(root->right, sum - root->val);    //获取右子节点的所有路径                for (int i = 0; i < rpaths.size(); ++i) {                    rpaths[i].insert(rpaths[i].begin(), root->val);                               //将当前节点插入到每条路径的起始位置                    paths.push_back(rpaths[i]);                }            }            return paths;        }};