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.
给出一颗二叉查找树，两个节点，找出这两个节点的最近公共祖先。
思路：递归可以使这道题变得异常的简单。如果两个节点是在根节点两侧，那么最近的公共祖先就是这个根节点，如果是左侧（右侧），则把根节点递归到左子树（右子树）。
/**
* 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;
}
int max = p.val > q.val ? p.val : q.val;
int min = p.val < q.val ? p.val : q.val;
if(max < root.val) {
return lowestCommonAncestor(root.left, p, q);
}
else if(min > root.val) {
return lowestCommonAncestor(root.right, p, q);
}
else {
return root;
} }
}