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]
]
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> ans = new LinkedList<List<Integer>>();
List<Integer> cur = new LinkedList<Integer>();
helper(root,sum,cur,ans);
return ans;
}
public void helper(TreeNode root,int sum,List<Integer> cur,List<List<Integer>> ans){
if(root == null) return;
cur.add(root.val);
if(root.left == null && root.right == null && root.val == sum){
ans.add(new LinkedList(cur));
cur.remove(cur.size() - 1);
return;
}else{
helper(root.left, sum - root.val, cur, ans);
helper(root.right, sum - root.val, cur, ans);
}
cur.remove(cur.size() - 1);
}