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700. Search in a Binary Search Tree ​

题目 ​

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
Related Topics
  • 树
  • 二叉搜索树
  • 二叉树

  • 👍 509
  • 👎 0

    • 思路 ​

    BST 的特征是，左侧 < 根 < 右侧

    解法一 ​

    python
    class Solution:
    def searchBST(self, root: TreeNode, val: int) -> TreeNode:
        if root is None:
            return None
        if val == root.val:
            return root
        return self.searchBST(root.left if val < root.val else root.right, val)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(N)

    解法二 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
# 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: Optional[TreeNode], val: int) -> Optional[TreeNode]:

        node = root

        while node:
            if node.val == val:
                return node
            elif node.val < val:
                node = node.right
            else:
                node = node.left
        return None


# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(1)