LeetCode Solutions : Path Sum I & II

Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:

Given the below binary tree and sum = 22,

5

/ \

4   8

/   / \

11  13  4

/  \      \

7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root==null)
			return false;
		if(root.left==null&&root.right==null)
			return sum==root.val?true:false;
		return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);
    }
}

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]

]

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        ArrayList<ArrayList<Integer>> results=new ArrayList<ArrayList<Integer>>();
		if(root==null)
			return results;
		helper(results,new ArrayList<Integer>(),root,sum);
		return results;
    }
    private void helper(ArrayList<ArrayList<Integer>> results,ArrayList<Integer> path,TreeNode root,int sum){
		if(root==null)
			return;
		path.add(root.val);
		if(root.left==null&&root.right==null){
			if(root.val==sum){
				ArrayList<Integer> temp=new ArrayList<Integer>(path);
				results.add(temp);
			}
		}
		helper(results,path,root.left,sum-root.val);
		helper(results,path,root.right,sum-root.val);
		path.remove(path.size()-1);
	}
}