leetcode-235. Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先

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.

给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。 百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。” 例如，给定如下二叉搜索树:  root = [6,2,8,0,4,7,9,null,null,3,5] 示例 1: 输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 输出: 6 解释: 节点 2 和节点 8 的最近公共祖先是 6。 示例 2: 输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 输出: 2 解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。 说明: 所有节点的值都是唯一的。 p、q 为不同节点且均存在于给定的二叉搜索树中。

思路：二叉搜索树有个性质 就是 左 子节点值<根节点值<右子节点值。那只需要判断 根节点和两个子节点个关系即可。若根节点值大于两个子节点值，那就说明两个节点是 根节点的左子节点，那就让根节点等于它的左子节点。若根节点小于两个子节点值，就说明两个子节点是根节点的右子节点，就让根节点指向 根节点右子节点。否则就说明大于 左子节点小于 右子节点。

递归方法：

/**
 * 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) return NULL;
        if(root->val > max(p->val,q->val))
            return lowestCommonAncestor(root->left,p,q);
        else if(root->val < min(p->val,q->val))
            return lowestCommonAncestor(root->right,p,q);
        else return root;
    }
};

 循环：

/**
 * 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) return NULL;
        while(true)
        {
            if(root->val > max(p->val,q->val))
                root = root->left;
            else if(root->val < min(p->val,q->val))
                root = root->right;
            else break;
        }
        return root;
    }
};