Leetcode NO.235 Lowest Common Ancestor of a Binary Search Tree

本题题目要求如下：

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.

本题乍一看不算简单，其实，只要考虑到BST的特性，还是可以比较容易的得到答案，

首先设定LCA为root，

(1)如果root->val比p,q值都小，LCA要设定为root->right

(2)如果root->val比p,q都大，LCA要设定为root->left

(3)如果root->val在p,q之间，则结束循环，LCA得到最终值：root

用recursive运行上面三步，如果遇到(3)跳出，返回最终值

代码如下：

/** * 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 (root->val < p->val and root->val < q->val) {            return lowestCommonAncestor(root->right, p, q);        }        else if (root->val > p->val and root->val > q->val) {            return lowestCommonAncestor(root->left, p, q);        }        else {            return root;        }    }};