129. Sum Root to Leaf Numbers

Hint

/**
 * 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:
    // Main function to initiate the DFS and return the total sum of all root-to-leaf numbers
    int sumNumbers(TreeNode* root) {
        
    }

    // Helper function to perform DFS traversal
    int dfs(TreeNode* root, int sum) {
        // If the current node is null, return 0 (base case for null nodes)
        
        // Update the sum by incorporating the current node's value
        
        
        // If the current node is a leaf (no children), return the computed sum
        
        // Recursively calculate the sum for left and right subtrees and return their sum
    }    
};

Solution - DFS

/**
 * 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 sumNumbers(TreeNode* root)
    {
        return dfs(root, 0);
    }
    int dfs(TreeNode* root, int sum)
    {
        if (!root) return 0;
        sum = sum * 10 + root->val;
        if (!root->left && !root->right) return sum;
        return dfs(root->left, sum) + dfs(root->right, sum);
    }    
};

• 時間複雜度：
  $O(n)$
• 空間複雜度：
  $O(H)$