《Cracking the Coding Interview》——第4章：树和图——题目5

2014-03-19 04:11

题目：设计算法检查一棵二叉树是否为二叉搜索树。

解法：既然是二叉搜索树，也就是说左子树所有节点都小于根，右子树所有节点都大于根。如果你真的全都检查的话，那就做了很多重复工作。只需要将左边最靠右，和右边最靠左的节点和根进行比较，然后依照这个规则递归求解即可。

代码：

 // 4.5 Check if a binary tree is binary search tree.
 #include <cstdio>
 using namespace std;

 struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;

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

 void constructBinaryTree(TreeNode *&root)
 {
     int val;

     scanf("%d", &val);
     if (val <= ) {
         root = nullptr;
     } else {
         root = new TreeNode(val);

         constructBinaryTree(root->left);
         constructBinaryTree(root->right);
     }
 }

 bool postorderTraversal(TreeNode *root, TreeNode *&left_most, TreeNode *&right_most)
 {
     TreeNode *ll, *lr, *rl, *rr;
     bool res_left = true, res_right = true;

     if (root->left != nullptr) {
         if (!postorderTraversal(root->left, ll, lr)) {
             return false;
         }
         if (lr->val >= root->val) {
             // all left nodes must be smaller than the root.
             return false;
         }
     } else {
         ll = lr = root;
     }

     if (root->right != nullptr) {
         if (!postorderTraversal(root->right, rl, rr)) {
             return false;
         }
         if (rl->val <= root->val) {
             // all right nodes must be greater than the root.
             return false;
         }
     } else {
         rl = rr = root;
     }
     left_most = ll;
     right_most = rr;

     return true;
 }

 void clearBinaryTree(TreeNode *&root) {
     if (root == nullptr) {
         return;
     } else {
         clearBinaryTree(root->left);
         clearBinaryTree(root->right);
         delete root;
         root = nullptr;
     }
 }

 int main()
 {
     TreeNode *root;
     TreeNode *left_most, *right_most;

     while (true) {
         constructBinaryTree(root);
         if (root == nullptr) {
             break;
         }

         left_most = right_most = nullptr;
         if (postorderTraversal(root, left_most, right_most)) {
             printf("Yes\n");
         } else {
             printf("No\n");
         }

         clearBinaryTree(root);
     }

     return ;
 }