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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

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• /**
   * 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);
      }
  }
  

• /**
   * 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;
          return root->val < 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.
   * 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)
  }
  

• /**
   * Definition for a binary tree node.
   * class TreeNode {
   *     val: number
   *     left: TreeNode | null
   *     right: TreeNode | null
   *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
   *         this.val = (val===undefined ? 0 : val)
   *         this.left = (left===undefined ? null : left)
   *         this.right = (right===undefined ? null : right)
   *     }
   * }
   */
  
  function searchBST(root: TreeNode | null, val: number): TreeNode | null {
      if (root === null || root.val === val) {
          return root;
      }
      return root.val > val ? searchBST(root.left, val) : searchBST(root.right, val);
  }
  
  

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