题目描述：
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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself
according to the LCA definition.
Note:
- All of the nodes' values will be unique.
- p and q are different and both values will exist in the BST.
寻找BST中两个节点的最近公共祖先，由于是BST，所以可以根据根节点的值来决定网哪边的子树搜索。一旦发现根节点的值位于两个目标值之间，那么说明它就是最近的公共祖先。
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
int low=min(p->val,q->val);
int high=max(p->val,q->val);
if(root->val==low||root->val==high) return root;
else if(root->val<high&&root->val>low) return root;
else if(root->val>high) return lowestCommonAncestor(root->left,p,q);
else if(root->val<low) return lowestCommonAncestor(root->right,p,q);
}
};