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],[5,8,5] ]
time: O(n),space: O(height)
/** * DeFinition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List<List<Integer>> pathSum(TreeNode root,int sum) { List<List<Integer>> res = new ArrayList<>(); List<Integer> list = new ArrayList<>(); pathSum(root,sum,list,res); return res; } public void pathSum(TreeNode root,int sum,List<Integer> list,List<List<Integer>> res) { if(root == null) { return; } if(root.left == null && root.right == null) { if(sum == root.val) { list.add(root.val); res.add(new ArrayList<>(list)); list.remove(list.size() - 1); } return; } list.add(root.val); pathSum(root.left,sum - root.val,res); pathSum(root.right,res); list.remove(list.size() - 1); } }
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