Leetcode 235. 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.

 

分析：

 

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。此时对root->left递归求解

 

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。对root->right递归求解

 

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。

 

c递归代码：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     struct TreeNode *left;
 6  *     struct TreeNode *right;
 7  * };
 8  */
 9 struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
10     if(root == NULL || p == NULL || q == NULL)
11             return NULL;
12     if(root->val > p->val && root->val > q->val){
13         return lowestCommonAncestor(root->left, p, q);
14     }
15     else if(root->val < p->val && root->val < q->val){
16         return lowestCommonAncestor(root->right, p, q);
17     }
18     else{
19          return root;
20     }
21 }

c非递归代码：

 1 struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
 2     if(root == NULL || p == NULL || q == NULL)
 3             return NULL;
 4     while(root != NULL){
 5         if(root->val > p->val && root->val > q->val)
 6             root = root->left;
 7         else if(root->val < p->val && root->val < q->val)
 8             root = root->right;
 9         else
10             break;
11     }
12     return root;
13 }

C++代码：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
13         if(root == NULL || p == NULL || q == NULL)
14             return NULL;
15         int min1 = min(p->val, q->val);
16         int max1 = max(p->val, q->val);
17         //if((root->val >= p->val && root->val <= q->val) || (root->val >= q->val && root->val <= p->val))
18          //   return root;
19         if(max1 < root->val){
20             return lowestCommonAncestor(root->left, p, q);
21         }
22         else if(min1 > root->val){
23             return lowestCommonAncestor(root->right, p, q);
24         }
25         else{
26             return root;
27         }
28     }
29 };

leetcode编译器运行，第一个程序用时最少，依次递增。

除了这种解法，也可以把它当做普通的二叉树做。

 

posted @ 2016-07-30 20:42  琴影  阅读(291)  评论(0)  编辑  收藏  举报