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