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Formatted question description: https://leetcode.ca/all/700.html

700. Search in a Binary Search Tree (Easy)

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node’s value equals the given value. Return the subtree rooted with that node. If such node doesn’t exist, you should return NULL.

For example, 

Given the tree:

    4
   / \
  2   7
 / \
1   3

And the value to search: 2

You should return this subtree:

  2     
 / \ 
1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

Solution 1.

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode searchBST(TreeNode root, int val) {
            TreeNode node = root;
            while (node != null) {
                if (node.val == val)
                    return node;
                else if (node.val > val)
                    node = node.left;
                else
                    node = node.right;
            }
            return null;
        }
    }
    
    ############
    
    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode() {}
     *     TreeNode(int val) { this.val = val; }
     *     TreeNode(int val, TreeNode left, TreeNode right) {
     *         this.val = val;
     *         this.left = left;
     *         this.right = right;
     *     }
     * }
     */
    class Solution {
        public TreeNode searchBST(TreeNode root, int val) {
            if (root == null || root.val == val) {
                return root;
            }
            return root.val < val ? searchBST(root.right, val) : searchBST(root.left, val);
        }
    }
    
  • // OJ: https://leetcode.com/problems/search-in-a-binary-search-tree/
    // Time: O(H)
    // Space: O(H)
    class Solution {
    public:
        TreeNode* searchBST(TreeNode* root, int val) {
            if (!root) return nullptr;
            if (root->val == val) return root;
            return val > root->val ? searchBST(root->right, val) : searchBST(root->left, val);
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode:
    #     def __init__(self, val=0, left=None, right=None):
    #         self.val = val
    #         self.left = left
    #         self.right = right
    class Solution:
        def searchBST(self, root: TreeNode, val: int) -> TreeNode:
            if root is None or root.val == val:
                return root
            return (
                self.searchBST(root.right, val)
                if root.val < val
                else self.searchBST(root.left, val)
            )
    
    ############
    
    # Definition for a binary tree node.
    # class TreeNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution:
        def searchBST(self, root, val):
            """
            :type root: TreeNode
            :type val: int
            :rtype: TreeNode
            """
            if not root:
                return None
            if root.val == val:
                return root
            elif root.val < val:
                return self.searchBST(root.right, val)
            else:
                return self.searchBST(root.left, val)
    
  • /**
     * Definition for a binary tree node.
     * type TreeNode struct {
     *     Val int
     *     Left *TreeNode
     *     Right *TreeNode
     * }
     */
    func searchBST(root *TreeNode, val int) *TreeNode {
    	if root == nil || root.Val == val {
    		return root
    	}
    	if root.Val < val {
    		return searchBST(root.Right, val)
    	}
    	return searchBST(root.Left, val)
    }
    

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