码迷,mamicode.com
首页 > 其他好文 > 详细

[LeetCode] Lowest Common Ancestor of a Binary Search Tree

时间:2015-07-24 18:23:43      阅读:106      评论:0      收藏:0      [点我收藏+]

标签:c++   leetcode   

Lowest Common Ancestor of a Binary Search Tree

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /                  ___2__          ___8__
   /      \        /         0      _4       7       9
         /           3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

解题思路:

本题题意是找到二分查找树的两个节点的最低公共祖先。根据二分查找树的特点,两个节点的最低公共祖先一定是大于等于较小节点,并且小于等于较大节点。倘若某个节点N的值都比给定的两个节点的值大,则公共节点一定在N的左孩子节点中。倘若某个节点N的值都比给定的两个节点的值小,则公共节点一定在N的右孩子节点中。

代码如下:

/**
 * 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) {
        int minNum = min(p->val, q->val);
        int maxNum = max(p->val, q->val);
        while(root!=NULL){
            if(minNum <= root->val && root->val <= maxNum)
            {
                return root;
            }else if(root->val < minNum){
                root = root->right;
            }else{
                root = root->left;
            }
        }
        return NULL;
    }
};
转载请注明:康瑞的部落 ? [LeetCode] Lowest Common Ancestor of a Binary Search Tree

版权声明:本文为博主原创文章,未经博主允许不得转载。

[LeetCode] Lowest Common Ancestor of a Binary Search Tree

标签:c++   leetcode   

原文地址:http://blog.csdn.net/kangrydotnet/article/details/47043733

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!