[leetcode] 235. Lowest Common Ancestor of a Binary Search Tree 解题报告

题目链接：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

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.

思路：因为是二叉搜索数，因此是中序有序的，对于一个结点来说有三种情况：

1. 如果结点值比p，q都大，说明公共最低公共祖先在此结点左边。

2. 如果结点值比p，q都小，说明公共最低公共祖先在此结点右边。

3. 如果结点值在p，q之间，说明这就是最低公共祖先。

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 || !p || !q) return NULL;
if(root->val>=p->val&& root->val<=q->val ||(root->val<=p->val && root->val>=q->val))
return root;
if(root->val > p->val && root->val >q->val)
return lowestCommonAncestor(root->left, p, q);
if(root->val < p->val && root->val <q->val)
return lowestCommonAncestor(root->right, p, q);
return NULL;
}
};