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]
]
这道题犯了一个严重的新手错误,在每次找到path中元素和 == sum的时候, 写成了res.add(path)。 可是因为backtracking, path最终还是会回到空List. 这样的话不管什么输出,输入都是类似[[],[], ...]这样的。 一定要记得res.add(new ArrayList<>(path)). 这样的话,之后不管path如何变动,是不会影响res里面加的path的。
/**
* 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> path = new ArrayList<>();
helper(root, sum, path, res);
return res;
}
private void helper(TreeNode root, int sum, List<Integer> path, List<List<Integer>> res){
if (root == null){
return;
}
path.add(root.val);
if (root.left == null && root.right == null){
int pathSum = 0;
for (int i = 0; i < path.size(); i++){
pathSum += path.get(i);
}
if (pathSum == sum){
res.add(new ArrayList<Integer>(path));
}
}
helper(root.left, sum, path, res);
helper(root.right, sum, path, res);
path.remove(path.size() - 1);
}
}