leetcode 235. Lowest Common Ancestor of a Binary Search Tree（python）

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).”

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.
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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.
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Example 3:

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

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Note:

The number of nodes in the tree is in the range [2, 10^5].
-10^9 <= Node.val <= 10^9
All Node.val are unique.
p != q
p and q will exist in the BST.
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解析

根据题意，给定一个二叉搜索树 (BST)，找到 BST 中两个给定节点的最低共同祖先 (LCA)。可以用最朴素的方法，找出从根节点到 p 的路径，也找出从根节点到 q 的路径，从左往右同时遍历两个路径的节点，找到第一次出现不同节点的前一个节点就是题目要找的 LCA 。因为题目中的限制条件很宽松，所以这种方法不会超时。

解答

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution(object):
    def __init__(self):
        self.paths = []
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        
        self.dfs(root, [], p)
        self.dfs(root, [], q)
        path_p = self.paths[0]
        path_q = self.paths[1]
        idx = 0
        while idx<min(len(path_p), len(path_q)) and path_p[idx].val==path_q[idx].val:
            idx += 1
        return path_p[idx-1]
    
    def dfs(self, root, path, t):
        if not root: return
        if root == t:
            path.append(root)
            self.paths.append(path)
            return 
        if root.left:
            self.dfs(root.left, path+[root], t)
        if root.right:
            self.dfs(root.right, path+[root], t)
            

        

           
        	      
		
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运行结果

Runtime: 96 ms, faster than 13.02% of Python online submissions for Lowest Common Ancestor of a Binary Search Tree.
Memory Usage: 21.6 MB, less than 26.58% of Python online submissions for Lowest Common Ancestor of a Binary Search Tree.
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解析

另外我们可以定义一个 dfs ，表示以某个节点为根节点的子树包含的 p 或者 q 的个数，可能为 0 表示没有包含 p 或者 q ，可能为 1 表示只包含了 p 或者 q ，可能为 2 表示都包含，因为递归是从下往上返回结果，当第一次出现 2 的时候并且 result 为空的时候，该节点为 LCA 。

解答

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution(object):
    def __init__(self):
        self.result = None
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        self.dfs(root, p, q)
        return self.result
    
    def dfs(self, root, p, q):
        if not root: return 0
        count  = self.dfs(root.left, p, q) + self.dfs(root.right, p, q) 
        if root == p or root == q: 
            count += 1
        if count == 2 and self.result==None:
            self.result = root
        return count
            

        
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运行结果

Runtime: 72 ms, faster than 55.76% of Python online submissions for Lowest Common Ancestor of a Binary Search Tree.
Memory Usage: 21.4 MB, less than 55.96% of Python online submissions for Lowest Common Ancestor of a Binary Search Tree.
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原题链接：leetcode.com/problems/lo…

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