【19】235. Lowest Common Ancestor of a Binary Search Tree

235. Lowest Common Ancestor of a Binary Search Tree

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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.

Solution 1: dfs 这是lowest ancestor of binary tree的做法，BST: left->val < root->val < right->val 可以利用这个

 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 == root || q == root) return root;
14         TreeNode* left = lowestCommonAncestor(root -> left, p, q);
15         TreeNode* right = lowestCommonAncestor(root -> right, p, q);
16         if(left && right) return root;
17         return left ? left : right;
18     }
19 };

Solution 2: 递归

 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 -> val < root -> val && q -> val > root -> val) return root;
14         //TreeNode* left = lowestCommonAncestor(root -> left, p, q);
15         //TreeNode* right = lowestCommonAncestor(root -> right, p, q);
16         if(!root) return NULL;
17         if(root -> val > max(p -> val, q -> val)){
18             return lowestCommonAncestor(root -> left, p, q);//如果root值同时大于他俩，则要往左找
19         }
20         else if(root -> val < min(p -> val, q -> val)){
21             return lowestCommonAncestor(root -> right, p, q);//如果root值同时xiao于他俩(min)，则要往right找
22         }
23         else return root;
24     }
25 };

 

Solution 3：while

同理，只不过root一直在走（更新）

 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) return NULL;
14         while(true){
15             if(root -> val > max(p -> val, q -> val)){
16                 root = root -> left;
17             }else if(root -> val < min(p -> val, q -> val)){
18                 root = root -> right;
19             }else{
20                 break;
21             }
22         }
23         return root;
24     }
25 };

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

posted @ 2017-02-07 07:36  会咬人的兔子  阅读(150)  评论(0)  编辑  收藏  举报