LeetCode 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

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.
题目链接
题意：

给一个二叉搜索树，找到给定两个节点的最低的公共祖先节点。

这题首先应考虑二叉搜索树的性质，对于每一个节点，左子节点小于父节点，右子节点大于父节点，所以最低的公共祖先节点一定符合p和q分布在其左支和右支上的性质，通过判断p，q的val与root的val大小，判断p，q在root的哪边分支上，然后递归或循环，直到root满足左右分支分别有一个p或q为止。

代码如下：

class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
while ((root->val - p->val) * (root->val - q->val) > 0) {
if (root->val < p->val && root->val < q->val)
root = root->right;
else if (root->val > p->val && root->val > q->val)
root = root->left;
}
return root;
}
};