【LeetCode】235. Lowest Common Ancestor of a Binary Search Tree 解题报告（Java & Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • 日期

[LeetCode]

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

Total Accepted: 67533 Total Submissions: 178900 Difficulty: Easy

题目描述

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).”

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

Example 1:

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:

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.

Note:

1. All of the nodes’ values will be unique.
2. p and q are different and both values will exist in the BST.

题目大意

在一个BST中，查找p和q节点的最小公共祖先。

解题方法

注意是BST，那么使用分而治之的策略，用递归来找到左边和右边的最低的公共祖先。

答案：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null||p==root||q==root){
            return root;
        }
        //devide
        TreeNode left=lowestCommonAncestor(root.left,p,q);
        TreeNode right=lowestCommonAncestor(root.right,p,q);

        //conquer
        if(left!=null&&right!=null){
            return root;
        }else if(left!=null){
            return left;
        }else{
            return right;
        }
    }
}

AC:11ms

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二刷，python

第一遍做这个题是两年前，现在用Python刷这个感觉特别简单。

因为BST本身的属性，所以比较节点的值和根节点的值的大小就知道下一步去哪里查找了。很简单，看代码。

递归版本：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if min(p.val, q.val) <= root.val and max(p.val, q.val) >= root.val:
            return root
        elif p.val < root.val and q.val < root.val:
            return self.lowestCommonAncestor(root.left, p, q)
        elif p.val > root.val and q.val > root.val:
            return self.lowestCommonAncestor(root.right, p, q)

循环版本：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        while root:
            if p.val < root.val and q.val < root.val:
                root = root.left
            elif p.val > root.val and q.val > root.val:
                root = root.right
            else:
                break
        return root

这个题是236. Lowest Common Ancestor of a Binary Tree的特例，所以可以直接使用236的代码就能通过。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if not root or root == p or root == q:
            return root
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)
        if left and right:
            return root
        return left if left else right

日期

2016/5/1 14:05:25
2018 年 6 月 22 日 ———— 这周的糟心事终于完了
2018 年 11 月 19 日 —— 周一又开始了