0100 - 0199
0129 - Sum Root to Leaf Numbers (Medium)
Problem Link
https://leetcode.com/problems/sum-root-to-leaf-numbers/
Problem Statement
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 -> 3represents the number123.
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.
Approach 1: DFS
static int fast_io = []() { std::ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); return 0; }();
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
__attribute__((disable_sanitizer_instrumentation))
static int sumNumbers(const TreeNode* root, int s = 0) {
if (!root) return 0;
s = s * 10 + root->val;
if (!root->left && !root->right) return s;
return sumNumbers(root->left, s) + sumNumbers(root->right, s);
}
};