LeetCode/CodingInterview/68_CommonParentInTree.py at master · chenjun0210/LeetCode · GitHub

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# -*- coding: utf-8 -*-
# @File    : 68_CommonParentInTree.py
# @Date    : 2020-04-12
# @Author  : tc
"""
面试题68 - I. 二叉搜索树的最近公共祖先
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]

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。
注意：本题与主站 235 题相同：https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

参考：https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree/solution/er-cha-sou-suo-shu-de-zui-jin-gong-gong-zu-xian--2/

"""
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':
        # Value of current node or parent node.
        parent_val = root.val

        # Value of p
        p_val = p.val

        # Value of q
        q_val = q.val

        # If both p and q are greater than parent
        if p_val > parent_val and q_val > parent_val:
            return self.lowestCommonAncestor(root.right, p, q)
        # If both p and q are lesser than parent
        elif p_val < parent_val and q_val < parent_val:
            return self.lowestCommonAncestor(root.left, p, q)
        # We have found the split point, i.e. the LCA node.
        else:
            return root


    def lowestCommonAncestor2(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        # Value of p
        p_val = p.val

        # Value of q
        q_val = q.val

        # Start from the root node of the tree
        node = root

        # Traverse the tree
        while node:

            # Value of current node or parent node.
            parent_val = node.val

            if p_val > parent_val and q_val > parent_val:
                # If both p and q are greater than parent
                node = node.right
            elif p_val < parent_val and q_val < parent_val:
                # If both p and q are lesser than parent
                node = node.left
            else:
                # We have found the split point, i.e. the LCA node.
                return node