Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
Note: A leaf is a node with no children.
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: Top down (find all paths which adds up is equal to target sum)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List> pathSum(TreeNode root, int sum) {
List> result = new ArrayList<> ();
if (root == null) {
return result;
}
List path = new ArrayList<> ();
pathSumHelper (root, path, result, sum);
return result;
}
private void pathSumHelper (TreeNode root, List path, List> result, int sum) {
if (root == null) {
return;
}
if (root.left == null && root.right == null && root.val == sum) {
path.add (root.val);
result.add (new ArrayList<> (path));
path.remove (path.size () - 1);
return;
} else if (root.left == null && root.right == null && root.val != sum) {
return;
}
int newSum = sum - root.val;
path.add (root.val);
pathSumHelper (root.left, path, result, newSum);
pathSumHelper (root.right, path, result, newSum);
path.remove (path.size () - 1);
}
}