【LeetCode】235. Lowest Common Ancestor of a Binary Search Tree 公共祖先

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.

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

基本题目公共祖先

/**
 * 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 == NULL|| root==p||root == q
        ||(root!=NULL&&root->val<p->val&&root->val>q->val)
        ||(root!=NULL&& root->val>p->val&&root->val<q->val))return root;

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