113. Path Sum II (Tree; DFS)

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]
]

 

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    vector<vector<int>> pathSum(TreeNode *root, int sum) {
        pathGroup.clear();
        if(!root) return pathGroup;
  
        target = sum;
        vector<int> path;
        preOrder(root,0,path);
        return pathGroup;
        
    }
    void preOrder(TreeNode* node,int sum,vector<int> path){
        sum = node->val + sum;
        path.push_back(node->val);
        if(node->left)
        {           
            preOrder(node->left,sum,path);
        }
       
        if(node->right)
        {           
            preOrder(node->right,sum,path);
        }
        
        if(!node->left && !node->right)
        {
            if(sum==target)
            {
                pathGroup.push_back(path);
            }        
        }
    }
private:
    int target;
    vector<vector<int>> pathGroup;
};