剑指 Offer 68 - I. 二叉搜索树的最近公共祖先

 

思路#

方法一：迭代#

/**
 * 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 != NULL) {
            if(p->val < root->val && q->val < root->val) {
                root = root->left;
            } else if(p->val > root->val && q->val > root->val) {
                root = root->right; 
            } else {
                break;
            }
        }

        return root;
    }
};

 

 

方法二：递归#

/**
 * 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->val < p->val && root->val < q->val)
            return lowestCommonAncestor(root->right, p, q);
        if(root->val > p->val && root->val > q->val)
            return lowestCommonAncestor(root->left, p, q);
        return root;
    }
};

 

 

参考#

面试题68 - I. 二叉搜索树的最近公共祖先（迭代 / 递归，清晰图解）