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.

由于这是一颗二叉搜索树，我们根据nodes v and w值的大小，能确定他们位于根节点的左边还是右边。如果两个节点都在根节点的左子树上，那么他们的LCA也肯定在根节点的左子树；如果两个节点都在根节点的右子树上，那么他们的LCA也肯定在根节点的右子树；如果两个节点分别位于根结的左右子树，那么根节点就是他们的LCA。否则的话直接返回根节点即可。代码如下：

 /**

  * 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 null;

         if(p.val<root.val && q.val<root.val){       //在根节点左边

             return lowestCommonAncestor(root.left, p, q);

         }

         else if(p.val>root.val && q.val>root.val){  //在根节点右边

             return lowestCommonAncestor(root.right, p, q);

         }

         else return root;                           //在根节点两侧，或者其中一个节点与根节点相等

     }

 }