leetCode235.LowestCommonAncestorofaBinarySearchTree二排序树问题
235. Lowest Common Ancestor of a Binary Search Tree
创新互联公司是一家网站设计制作、成都网站建设,提供网页设计,网站设计,网站制作,建网站,专业公司,网站开发公司,从2013年创立是互联行业建设者,服务者。以提升客户品牌价值为核心业务,全程参与项目的网站策划设计制作,前端开发,后台程序制作以及后期项目运营并提出专业建议和思路。
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.
思路:
mine思路:
1.将各个节点的从root到节点的路径都记录下来vector
2.找到目标的两个vector,比较之找到较短的"根"
others思路:
在二叉查找树种,寻找两个节点的最低公共祖先。
1、如果a、b都比根节点小,则在左子树中递归查找公共节点。
2、如果a、b都比根节点大,则在右子树中查找公共祖先节点。
3、如果a、b一个比根节点大,一个比根节点小,或者有一个等于根节点,则根节点即为最低公共祖先。
代码如下:(代码实现为others思路)
/** * 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 pVal = p->val; int qVal = q->val; int rootVal = root->val; if(pVal < rootVal && qVal < rootVal) { return lowestCommonAncestor(root->left,p,q); } else if(pVal > rootVal && qVal > rootVal) { return lowestCommonAncestor(root->right,p,q); } return root; } };
2016-08-07 09:47:25
网页题目:leetCode235.LowestCommonAncestorofaBinarySearchTree二排序树问题
地址分享:http://azwzsj.com/article/pcphci.html