# [Leetcode 700] Search in a Binary Search Tree

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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 = 2Output: [2,1,3]`

Example 2:

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

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;}`