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Java-based LeetCode algorithm problem solutions, regularly updated
package g0101_0200.s0129_sum_root_to_leaf_numbers;
// #Medium #Depth_First_Search #Tree #Binary_Tree
// #2022_06_23_Time_0_ms_(100.00%)_Space_41.8_MB_(46.81%)
import com_github_leetcode.TreeNode;
/*
* 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;
* }
* }
*/
/**
* 129 - Sum Root to Leaf Numbers\.
*
* Medium
*
* You are given the `root` of a binary tree containing digits from `0` to `9` only.
*
* Each root-to-leaf path in the tree represents a number.
*
* * For example, the root-to-leaf path `1 -> 2 -> 3` represents the number `123`.
*
* Return _the total sum of all root-to-leaf numbers_. Test cases are generated so that the answer will fit in a **32-bit** integer.
*
* A **leaf** node is a node with no children.
*
* **Example 1:**
*
* ![](https://assets.leetcode.com/uploads/2021/02/19/num1tree.jpg)
*
* **Input:** root = [1,2,3]
*
* **Output:** 25
*
* **Explanation:** The root-to-leaf path `1->2` represents the number `12`. The root-to-leaf path `1->3` represents the number `13`. Therefore, sum = 12 + 13 = `25`.
*
* **Example 2:**
*
* ![](https://assets.leetcode.com/uploads/2021/02/19/num2tree.jpg)
*
* **Input:** root = [4,9,0,5,1]
*
* **Output:** 1026
*
* **Explanation:** The root-to-leaf path `4->9->5` represents the number 495. The root-to-leaf path `4->9->1` represents the number 491. The root-to-leaf path `4->0` represents the number 40. Therefore, sum = 495 + 491 + 40 = `1026`.
*
* **Constraints:**
*
* * The number of nodes in the tree is in the range `[1, 1000]`.
* * `0 <= Node.val <= 9`
* * The depth of the tree will not exceed `10`.
**/
public class Solution {
private int sum = 0;
public int sumNumbers(TreeNode root) {
recurseSum(root, 0);
return sum;
}
private void recurseSum(TreeNode node, int curNum) {
if (node.left == null && node.right == null) {
sum += 10 * curNum + node.val;
} else {
if (node.left != null) {
recurseSum(node.left, 10 * curNum + node.val);
}
if (node.right != null) {
recurseSum(node.right, 10 * curNum + node.val);
}
}
}
}