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 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.
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
