[Leetcode 700] Search in a Binary Search Tree

Description:

You are given the root of a binary search tree (BST) and an integer val.

Find the node in the BST that the node’s value equals val and return the subtree rooted with that node. If such a node does not exist, return null.

Example 1:

Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]

Example 2:

Input: root = [4,2,7,1,3], val = 5
Output: []

Constraints:

• The number of nodes in the tree is in the range [1, 5000].
• 1 <= Node.val <= 107
• root is a binary search tree.
• 1 <= val <= 107

C code solution (recursive):

struct TreeNode* searchBST(struct TreeNode* root, int val){
    //boundary
    if(!root)
        return NULL;
    
    //main
    struct TreeNode* ptr = root;
    if(val < ptr->val)
        ptr = searchBST(ptr->left, val);
    else if(val > ptr->val)
        ptr = searchBST(ptr->right, val);
    else
        return ptr;
    
    return ptr;
}

C code solution (iterative):

struct TreeNode* searchBST(struct TreeNode* root, int val){
    
    struct TreeNode* ptr = root;
    while(ptr){
        if(val < ptr->val)
            ptr = ptr->left;
        else if(val > ptr->val)
            ptr = ptr->right;
        else
            return ptr;
    }
    
    return ptr;
}