【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.

提示：

此题的核心是要利用已知条件中，给定的二叉树是二叉搜索树这一条件。所谓二叉搜索树，指的是一种二叉树，其中，比根节点小的节点都位于根节点的左侧，反之，若比根节点要大，则位于根节点的右侧。可以参考题目中的例子。

递归调用的思路也很简单：

• 若两个要搜索的节点都比根节点小，则对根节点的左节点进行递归；
• 若两个要搜索的节点都比根节点打，则对根节点的右节点进行递归；
• 若一大一小，则返回根节点。

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);
        if (p->val > root->val && q->val > root->val)
            return lowestCommonAncestor(root->right, p, q);
        return root;
    }
};

由于此题的测试用例中没有节点为NULL的情况，因此代码中省去了空指针的判断。