Careercup - Facebook面试题 - 5188884744896512

2014-05-02 07:18

题目链接

原题：

boolean isBST(const Node* node) { 

// return true iff the tree with root 'node' is a binary search tree. 

// 'node' is guaranteed to be a binary tree. 

} 



  n

 / \

a   b

     \

      c

题目：检查一棵二叉树是否为二叉搜索树。

解法：二叉搜索树，就是每个节点的左边全都小于它，右边全都大于它。如果真的对于每个节点都全部检查左右边的每个节点，就做了很多重复劳动。只需要判断每个节点的左子树最靠右，和右子树最靠左的节点是否小于和大于它即可。递归过程中传递引用可以随时更新两个需要检查的值。请看代码。

代码：

  1 // http://www.careercup.com/question?id=5632735657852928

  2 #include <climits>

  3 #include <iostream>

  4 #include <sstream>

  5 #include <string>

  6 using namespace std;

  7 

  8 struct TreeNode {

  9     int val;

 10     TreeNode *left;

 11     TreeNode *right;

 12     TreeNode(int _val = 0): val(_val), left(nullptr),right(nullptr) {};

 13 };

 14 

 15 class Solution {

 16 public:

 17     bool isBST(TreeNode *root) {

 18         if (root == nullptr) {

 19             return false;

 20         }

 21         

 22         max_val = INT_MIN;

 23         res = true;

 24         first_node = true;

 25         isBSTRecursive(root);

 26         

 27         return res;

 28     };

 29 private:

 30     int max_val;

 31     bool res;

 32     bool first_node;

 33     

 34     void isBSTRecursive(TreeNode *root) {

 35         if (!res) {

 36             return;

 37         }

 38         

 39         // root is guaranteed to be not nullptr.

 40         if (root->left) {

 41             isBSTRecursive(root->left);

 42         }

 43         if (first_node || root->val > max_val) {

 44             first_node = false;

 45             max_val = root->val;

 46         } else {

 47             res = false;

 48             return;

 49         }

 50         if (root->right) {

 51             isBSTRecursive(root->right);

 52         }

 53     };

 54 };

 55 

 56 void construcTree(TreeNode *&root)

 57 {

 58     int val;

 59     stringstream sio;

 60     string s;

 61     

 62     if (cin >> s && s != "#") {

 63         sio << s;

 64         sio >> val;

 65         root = new TreeNode(val);

 66         construcTree(root->left);

 67         construcTree(root->right);

 68     } else {

 69         root = nullptr;

 70     }

 71 }

 72 

 73 void deleteTree(TreeNode *&root)

 74 {

 75     if (root == nullptr) {

 76         return;

 77     }

 78     deleteTree(root->left);

 79     deleteTree(root->right);

 80     delete root;

 81     root = nullptr;

 82 }

 83 

 84 int main()

 85 {

 86     TreeNode *root;

 87     Solution sol;

 88     

 89     while (true) {

 90         construcTree(root);

 91         if (root == nullptr) {

 92             break;

 93         }

 94         

 95         cout << (sol.isBST(root) ? "Valid BST" : "Invalid BST") << endl;

 96         

 97         deleteTree(root);

 98     }

 99     

100     return 0;

101 }