LeetCode //C - 129. Sum Root to Leaf Numbers

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.

From: LeetCode
Link: 129. Sum Root to Leaf Numbers

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Solution:

Ideas：

1. Use a recursive helper function to traverse the tree.

2. In the helper function:

• If the current node is null, return 0.
• Compute the current number by multiplying the parent’s number by 10 and adding the current node’s value.
• If the current node is a leaf node (both left and right children are null), return the current number.
• Otherwise, return the sum of the helper function’s results for the left and right children.

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int helper(struct TreeNode* root, int currentNumber) {
    if (!root) {
        return 0;
    }

    // Compute the current number
    currentNumber = currentNumber * 10 + root->val;

    // If it's a leaf node
    if (!root->left && !root->right) {
        return currentNumber;
    }

    return helper(root->left, currentNumber) + helper(root->right, currentNumber);
}

int sumNumbers(struct TreeNode* root) {
    return helper(root, 0);
}
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