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]]

 

 

Code:

/** * 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 findPath(TreeNode *node, int n, int &sum, vector
     
       &buf, vector
      
       
        > &res){        if(!node)            return;        buf.push_back(node->val);        if(!node->left&&!node->right&&node->val+n==sum){            res.push_back(buf);            buf.pop_back();            return;        }        findPath(node->left,node->val+n,sum,buf,res);        findPath(node->right,node->val+n,sum,buf,res);        buf.pop_back();    }    vector
        
         
           > pathSum(TreeNode *root, int sum) {            vector
          
            buf; vector
           
            
             > res; findPath(root,0,sum,buf,res); return res; }};