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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:

Ex1

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

Example 2:

Ex2

  • 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 <= 10^7
  • root is a binary search tree.
  • 1 <= val <= 10^7

Solution⚓︎

Binary Search Tree is a binary tree where the key in each node:

  • is greater than any key stored in the left sub-tree,
  • and less than any key stored in the right sub-tree.

Recursive Solution⚓︎

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if (!root || root->val == val) return root;
        if (root->val > val) return searchBST(root->left, val);
        if (root->val < val) return searchBST(root->right, val);
        return nullptr;
    }
};

Another way of writing:

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class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if (!root || root->val == val) return root;
        TreeNode *res = nullptr;
        if (root->val > val) res = searchBST(root->left, val);
        if (root->val < val) res = searchBST(root->right, val);
        return res;
    }
};
  • Time complexity (worst): \(O(n)\);
  • Space complexity (worst): \(O(n)\) for recursion stack.

Iterative Solution⚓︎

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class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        TreeNode* curr = root;
        while (curr && curr->val != val) {
            if (curr->val > val) curr = curr->left;
            else curr = curr->right;
        }
        return curr;
    }
};
  • Time complexity (worst): \(O(n)\);
  • Space complexity: \(O(1)\).