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 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]

Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes `2` and `8` is `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.

解析

二叉查找树（Binary Search Tree），（又：二叉搜索树，二叉排序树）它或者是一棵空树，或者是具有下列性质的二叉树： 若它的左子树不空，则左子树上所有结点的值均小于它的根结点的值； 若它的右子树不空，则右子树上所有结点的值均大于它的根结点的值； 它的左、右子树也分别为二叉排序树。
以上是二叉查找树的定义，可以发现其前序遍历是由小到大的排列。因此，
对于求公共祖先，可以依据以下事实：
（1）如果p,q的最大值都比根结点要小，那么其公共祖先在左子树上；
（2）如果p,q的最小值都比根结点要大，那么其公共祖先在右子树上；
（3）如果p和q分居在root的左右两侧，那么其公共祖先为根结点。

其它结点进行递归求解。

代码（C语言）

struct TreeNode* lowestCommonAncestor(struct TreeNode* root,
                                      struct TreeNode* p, struct TreeNode* q) {
    if (root == NULL)
        return NULL;
    
    int minNum = p->val > q->val ? q->val : p->val;
    int maxNum = p->val > q->val ? p->val : q->val;
    
    if (minNum > root->val) {
        return lowestCommonAncestor(root->right, p, q);
    } else if (maxNum < root->val) {
        return lowestCommonAncestor(root->left, p, q);
    }
    
    return root;
}