[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;
}

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