题目描述

原题链接：235. 二叉搜索树的最近公共祖先

解题思路

1、普通树找公共祖先解法

/**
 * 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 || root == p || root ==q)      return root;
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        if(left && right)       return root;
        if(left && !right)      return left;
        return right;
    }
};

2、利用BST性质解法

递归法

/**
 * 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;
        // 当p和q都在左侧时，向左遍历
        if(root->val > p->val && root->val > q->val) {
            return lowestCommonAncestor(root->left, p, q);
        }
        // 当p和q都在右侧时，向右遍历
        if(root->val < p->val && root->val < q->val) {
            return lowestCommonAncestor(root->right, p, q);
        }
        // 当p和q分别在两侧时，第一个找到的这种结点一定为LCA
        return root;
    }
};

迭代法

/**
 * 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) {
        while(root) {
            int value = root->val;
            if(value > p->val && value > q->val)     root = root->left;
            else if(value < p->val && value < q->val)   root = root->right;
            else    return root;
        }
        return NULL;
    }
};

参考文章：235. 二叉搜索树的最近公共祖先