235. Lowest Common Ancestor of a Binary Search Tree

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

​​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：

本题题目出得花里胡哨的，其实这类BTS的题目都是毫无意义的，只要抓住递归的本质，首先做这类与数据结构密切相关的题目时，一定要熟悉BTS的性质：

（1）若左子树不空，则左子树上所有结点的值均小于或等于它的​​根结​​点的值；

（2）若右子树不空，则右子树上所有结点的值均大于或等于它的根结点的值；

（3）左、右子树也分别为二叉排序树；

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(p->val<root->val&&q->val<root->val)
            return lowestCommonAncestor(root->left,p,q);
        else if(p->val>root->val&&q->val>root->val)
            return lowestCommonAncestor(root->right,p,q);
        else
            return root;
        
    }
};