129. Sum Root to Leaf Numbers

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

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.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Tree Traversal (Duyệt cây)String (Chuỗi)
⏱️ Thời gian O(V+E)
💾 Không gian O(V)

💻 Lời Giải

C++ 0129-sum-root-to-leaf-numbers.cpp 
/**
 * 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:
    int dfs(TreeNode *root, string num) {
        if (!root) {
            return 0;
        }
        if (!root->left and !root->right) {
            return stoi(num + char(root->val + '0'));
        }
        int cnt = 0;
        cnt += dfs(root->left, num + char(root->val + '0'));
        cnt += dfs(root->right, num + char(root->val + '0'));
        return cnt;
    }
    int sumNumbers(TreeNode* root) {
        return dfs(root, "");
    }
};