java 二叉搜索树寻找最小跟结点

题目：
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.

如图，根据二叉搜索树的特征，比根结点小的数位于二叉搜索数左子树上，比根结点大的数位于二叉搜索树的右子树上。所以如果两个结点分别位于左右子树，则此根结点就是所求。如果两结点最大的值小于根结点值，递归左子树，如果两结点最小的值大于根结点值，递归右子树。算法如下：

/** * 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(Math.max(p.val,q.val)<root.val){            root = root.left;            return lowestCommonAncestor(root,p,q);        }else if(Math.min(p.val,q.val)>root.val){            root = root.right;            return lowestCommonAncestor(root,p,q);        }        return root;    }}