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]
]





Solution :

The idea is to use DFS. If in a path from root to a leaf the sum is found, the path

is entered into a result vector. If not, we backtrack and look for other paths.



Code :

	class Solution {
	public:
	vector<vector<int> > pathSum(TreeNode *root, int sum) {
	vector<vector<int>> vec;
	vector<int> ivec;
	findSumPath (root, sum, vec, ivec);
	return vec;
	}
	public:
	void findSumPath (TreeNode *root, int sum, vector<vector<int>> &vec, vector<int> &ivec) {
	if(!root) return;
	ivec.push_back(root->val);
	if(root->left == NULL && root->right== NULL && sum == root->val)
	{
	vec.push_back(ivec);
	ivec.pop_back();
	return;
	}
	if (root->left) findSumPath(root->left, sum-root->val, vec, ivec);
	if (root->right) findSumPath(root->right, sum-root->val, vec, ivec);
	ivec.pop_back();
	}
	};

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