Day 21, 113, Path Sum II

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]
] 

---------- 

based on Path Sum

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void ps (TreeNode *root, int sum, int curSum, vector<int> curV, vector<vector<int> > &re) {
        if (root != NULL) {
            curSum += root->val;
            curV.push_back(root->val);
            if (root->left == NULL && root->right == NULL) {
                if (curSum == sum) {
                    re.push_back(curV);
                }
            }else {
                ps(root->left,sum,curSum,curV,re);
                ps(root->right,sum,curSum,curV,re);
            }
        }
    }

    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> path;
        vector<vector<int> > paths;
        ps(root,sum,0,path,paths);
        return paths;
    }
};

优化了下用于暂时储存的vector，此法可用于若干种类似题目

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void helper(TreeNode *root,int sum, vector<vector<int> > &rt, vector<int> &cur) {
        if (root == NULL) return;
        if (root->val == sum && root->left == NULL && root->right == NULL) {
            vector<int> t = cur;
            t.push_back(root->val);
            rt.push_back(t);
            return;
        }
        
        cur.push_back(root->val);
        helper(root->left,sum - root->val,rt,cur);
        helper(root->right,sum - root->val,rt,cur);
        cur.pop_back();
    }

    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int> > rt;
        vector<int> cur;
        
        helper(root,sum,rt,cur);
        return rt;
    }
};