LeetCode | 0700. Search in a Binary Search Tree二叉搜索樹中的搜索【Python】

LeetCode 0700. Search in a Binary Search Tree二叉搜索樹中的搜索【Easy】【Python】【二叉樹】

Problem

LeetCode

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.

問題

力扣

給定二叉搜索樹（BST）的根節點和一個值。 你需要在BST中找到節點值等於給定值的節點。 返回以該節點爲根的子樹。 如果節點不存在，則返回 NULL。

例如，

給定二叉搜索樹:

    4
   / \
  2   7
 / \
1   3

和值: 2
你應該返回如下子樹:

  2     
 / \   
1   3

在上述示例中，如果要找的值是 5，但因爲沒有節點值爲 5，我們應該返回 NULL。

思路

二叉樹

Python3代碼

# 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: TreeNode, val: int) -> TreeNode:
        # 節點不存在
        if not root:
            return None
        if root.val == val:
            return root
        if root.val < val:
            return self.searchBST(root.right, val)
        if root.val > val:
            return self.searchBST(root.left, val)

GitHub鏈接

Python