18. Sum of leaf nodes in BST

The problem can be found at the following link: Question Link

My Approach

• If the root is null, it returns 0.

• If the root has no left and right children (i.e., it's a leaf node), it returns the value of that node.

• Otherwise, it recursively calls itself on the left and right subtrees and returns the sum of the results.

Time and Auxiliary Space Complexity

• Time Complexity : O(N), where N is the number of nodes in the Binary Search Tree.

• Auxiliary Space Complexity : O(H), where H is the height of the Binary Search Tree.

Code (C++)

class Solution {
public:
    int sumOfLeafNodes(Node *root) {
        if (!root)
            return 0;
        
        if (!root->left && !root->right)
            return root->data;
        
        return sumOfLeafNodes(root->left) + sumOfLeafNodes(root->right);
    }
};

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