LC.235.Lowest Common Ancestor of a Binary Search Tree

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/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         /  \         3   5For 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.像这种返回一个node 的东西，结果要不就从左边来， 要不就从右边来，否则就是null凡事遍历的点，先查一下自己满足不满足，不满足的话，稍微pruning 一下，大胆往左右两边丢，然后等答案返回来check bal. tree height, path sum i 都是这样

 1 // the goal is to find the root that would sit in the middle
 2     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
 3         //base case
 4         if (root == null) return null ;
 5         int min = Math.min(p.val, q.val) ;
 6         int max = Math.max(p.val, q.val) ;
 7         //pre order：看看当前点是不是满足的值
 8         if (root.val>=min && root.val <= max) return root ;
 9         //the left: branch pruning > max, then go left： 左边返回就一层层往上返回
10         if (root.val > max){
11            return lowestCommonAncestor(root.left, p, q) ;
12         }
13         //the right: branch pruning: <min: then go right： 右边返回就一层层往上返回
14         if (root.val < min){
15           return lowestCommonAncestor(root.right, p, q) ;
16         }
17         //不从左边，也不从右边的话，就是NULL
18         return null;
19     }

这道题首先要明确题目背景---二叉搜索树。也正是因为是二叉搜索树，所以我们可以利用二叉搜索树从小到大排好序的特性来做。

对于一个root和另外两个Node来说，它们的值会有以下几种情况：

1. root.val < p.val && root.val < q.val 此时，两个node的值都比root大，说明这两个值在root右边的subtree里，且他们的最小公共顶点不可能是当前root

2. root.val > p.val && root.val > q.val 此时，两个node的值都比root小，说明这两个值在root左边的subtree里，且他们的最小公共顶点不可能是当前root

3. else,此时 root的值在p，q之间（或等于p，q中的某一个），此时，当前root即为公共顶点

原文地址：https://www.cnblogs.com/davidnyc/p/8495924.html