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;
}

}