题目描述：
一个二叉搜索树，给定两个节点a，b，求最小的公共祖先
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5例如：
2,8 —->6 2,4—–>2
原文描述：
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
/ \
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.
思路分析：
- 二叉树考虑递归的思路
- 如果a < = root <= b，可以确定root是lct
- 采用二分搜索的方式递归，root > a,b，继续遍历左子树，反之右边的子树
-在递归的开始判断空值的情况，直接返回root
代码：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root == null || p == null || q == null)
return root;
if(p.val > root.val && q.val > root.val)
return lowestCommonAncestor(root.right, p, q);
else if(p.val < root.val && q.val < root.val)
return lowestCommonAncestor(root.left, p, q);
else
return root;
}
}