Leetcode 235.二叉搜索树的最近公共祖先

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235. 二叉搜索树的最近公共祖先

题目

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）。”

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

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。

方法

方法1：递归

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# 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, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        p_val=p.val
        q_val=q.val
        r_val=root.val

        if p_val>r_val and q_val>r_val:
            return self.lowestCommonAncestor(root.right,p,q)
        elif p_val<r_val and q_val<r_val:
            return self.lowestCommonAncestor(root.left,p,q)
        else:
            return root

Complexity Analysis

  • Time Complexity: O(N), where N is the number of nodes in the BST. In the worst case we might be visiting all the nodes of the BST.
  • Space Complexity: O(N). This is because the maximum amount of space utilized by the recursion stack would be N since the height of a skewed BST could be N.

方法2：非递归

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# 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, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        p_val=p.val
        q_val=q.val

        node=root
        while node:
            r_val=node.val

            if p_val<r_val and q_val<r_val:
                node=node.left
            elif p_val>r_val and q_val>r_val:
                node=node.right
            else:
                return node

Complexity Analysis

  • Time Complexity : O(N), where N is the number of nodes in the BST. In the worst case we might be visiting all the nodes of the BST.
  • Space Complexity : O(1).
  • Leetcode Solution