LeetCode-in-Java
113. Path Sum II
Medium
Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.
A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum: 5 + 4 + 11 + 2 = 22 5 + 8 + 4 + 5 = 22
Example 2:
Input: root = [1,2,3], targetSum = 5
Output: []
Example 3:
Input: root = [1,2], targetSum = 0
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]. -1000 <= Node.val <= 1000-1000 <= targetSum <= 1000
Solution
import com_github_leetcode.TreeNode;
import java.util.ArrayList;
import java.util.List;
/*
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
recur(res, new ArrayList<>(), 0, targetSum, root);
return res;
}
private void recur(
List<List<Integer>> res, ArrayList<Integer> al, int sum, int targetSum, TreeNode root) {
if (root == null) {
return;
}
al.add(root.val);
sum += root.val;
if (sum == targetSum && root.left == null && root.right == null) {
res.add(new ArrayList<>(al));
}
recur(res, al, sum, targetSum, root.left);
recur(res, al, sum, targetSum, root.right);
al.remove(al.size() - 1);
}
}