[LeetCode] 129. Sum Root to Leaf Numbers

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:

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:

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.

求根到叶子节点数字之和。

给你一个二叉树的根节点 root ，树中每个节点都存放有一个 0 到 9 之间的数字。
每条从根节点到叶节点的路径都代表一个数字：

例如，从根节点到叶节点的路径 1 -> 2 -> 3 表示数字 123 。
计算从根节点到叶节点生成的所有数字之和。

叶节点是指没有子节点的节点。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/sum-root-to-leaf-numbers
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是前序遍历。记录一个变量 sum 存之前所有的加和，当遍历到当前节点的时候，sum *= 10 再加当前的节点值 cur.val。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     int res = 0;
18 
19     public int sumNumbers(TreeNode root) {
20         helper(root, 0);
21         return res;    
22     }
23 
24     private void helper(TreeNode root, int sum) {
25         if (root == null) {
26             return;
27         }
28         sum = sum * 10 + root.val;
29         if (root.left == null && root.right == null) {
30             res += sum;
31             return;
32         }
33         helper(root.left, sum);
34         helper(root.right, sum);
35     }
36 }

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @return {number}
 4  */
 5 var sumNumbers = function (root) {
 6     if (root == null) {
 7         return 0;
 8     }
 9     var total = 0;
10     helper(root, 0);
11     return total;
12 
13     function helper(root, sum) {
14         if (root == null) {
15             return;
16         }
17         sum = sum * 10 + root.val;
18         if (root.left == null && root.right == null) {
19             total += sum;
20             return;
21         }
22         helper(root.left, sum);
23         helper(root.right, sum);
24     }
25 };

CNoodle

Let the code speak for you.

[LeetCode] 129. Sum Root to Leaf Numbers

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