Question

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 = 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

Solution 1

# 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.
 * 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.
 * 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)
}