[LeetCode]Lowest Common Ancestor of a Binary Search Tree

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.

用到二叉搜索树本问题就很好解。

二叉搜索树的特点就是root->leff < root < root->right。

如果root>p && root>q，那么LCA在左子树上；

如果root<p && root<q，那么LCA在右子树上。

递归实现。

 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 || !p || !q) return NULL;
14         if(root->val>p->val && root->val>q->val) return lowestCommonAncestor(root->left,p,q);
15         if(root->val<p->val && root->val<q->val) return lowestCommonAncestor(root->right,p,q);
16         return root;
17     }
18 };

迭代实现。

 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 || !p || !q) return NULL;
14         TreeNode *cur = root;
15         while(cur)
16         {
17             if(cur->val>p->val && cur->val>q->val) cur = cur->left;
18             else if(cur->val<p->val && cur->val<q->val) cur = cur->right;
19             else return cur;
20         }
21         return cur;
22     }
23 };