LeetCode#235 Lowest Common Ancestor of a Binary Search Tree

问题描述

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

[6, 2, 8, 0, 4, 7, 9, null, null, 3, 5]

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

补充说明：

给定了一个二叉搜索树，输入两个值，然后求这两个值得最近公共祖先。

方案分析

1. 笔者做这个题的时候忘记了一个问题，导致感觉这个题写起来比较复杂。后来参照了别人的解法才恍然大悟。这里都说了是一个二叉搜索树，那么，这里隐藏的一个条件是搜索树是有序的！
2. 父节点的值肯定是大于左孩子节点的值，小于右边孩子节点的值。那么给定两个要搜索的值，它的最近共同祖先的要不就是介于两个值之间，要不就是等于较小值。（等于是当两个搜索值在同侧的情况下，其中的一个搜索值作为自己本身的祖先）。
3. 先上基本解法的递归写法。

python实现

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

class Solution:

    def lowestCommonAncestor(self, root, p, q):
        if not root:
            return 
    
        if root.val > p.val and root.val > q.val:
            return self.lowestCommonAncestor(root.left, p, q)
        elif root.val < p.val and root.val < q.val:
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return root

方案分析2

1. 递归并不是一个好的方法，换非递归的。

python实现2

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

class Solution:

    def lowestCommonAncestor(self, root, p, q):
        while root:
            if root.val > p.val and root.val > q.val:
                root = root.left
            elif root.val < p.val and root.val < q.val:
                root = root.right
            else:
                return root

方案分析3

1. 在别人那看到一个应用数学特性解决问题的方式，效率提升明显。
2. 如果一个值x的大小介于a,b之间，那么(x-a)*(x-b)<=0。这不正是二叉搜索树的特性么。

python实现3

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

class Solution:

    def lowestCommonAncestor(self, root, p, q):
        while (root.val-p.val)*(root.val-q.val) > 0:
            if root.val > max(p.val, q.val):
                root = root.left
            else:
                root = root.right
        return root