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LeetCode | 0235. 二叉搜索树的最近公共祖先【Py

LeetCode | 0235. 二叉搜索树的最近公共祖先【Py

作者: Wonz | 来源:发表于2021-01-25 21:08 被阅读0次

    Problem

    LeetCode

    Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

    According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

    Example 1:

    [外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-dGzMQij7-1610372086960)(https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)]

    Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
    Output: 6
    Explanation: The LCA of nodes 2 and 8 is 6.
    

    Example 2:

    [外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-KanePah3-1610372086969)(https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png)]

    Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
    Output: 2
    Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
    

    Example 3:

    Input: root = [2,1], p = 2, q = 1
    Output: 2
    

    Constraints:

    • The number of nodes in the tree is in the range [2, 105].
    • -109 <= Node.val <= 109
    • All Node.val are unique.
    • p != q
    • p and q will exist in the BST.

    问题

    力扣

    给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。

    百度百科中最近公共祖先的定义为:“对于有根树 T 的两个结点 p、q,最近公共祖先表示为一个结点 x,满足 x 是 p、q 的祖先且 x 的深度尽可能大(一个节点也可以是它自己的祖先)。”

    例如,给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]

    img

    示例 1:

    输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
    输出: 6 
    解释: 节点 2 和节点 8 的最近公共祖先是 6。
    

    示例 2:

    输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
    输出: 2
    解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。
    

    说明:

    • 所有节点的值都是唯一的。
    • p、q 为不同节点且均存在于给定的二叉搜索树中。

    思路

    遍历

    利用二叉搜索树的性质,中序遍历就是递增的顺序。
    于是,可以同时判断 p、q 大小,看看都在左边还是都在右边还是左右都有。
    

    Python3 代码

    # Definition for a binary tree node.
    # class TreeNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution:
        def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
            ancestor = root
            while True:
                # p q 在左边
                if p.val < ancestor.val and q.val < ancestor.val:
                    ancestor = ancestor.left
                # p q 在右边
                elif p.val > ancestor.val and q.val > ancestor.val:
                    ancestor = ancestor.right
                # p q 分别在左右,当前节点就是最近公共祖先
                else:
                    break
            return ancestor
    

    GitHub 链接

    Python

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