《Cracking the Coding Interview》——第17章：普通题——题目13

2014-04-29 00:15

题目：将二叉搜索树展开成一个双向链表，要求这个链表仍是有序的，而且不能另外分配对象，就地完成。

解法：Leetcode上也有，递归解法。

代码：

 // 17.13 Flatten a binary search tree into a doubly linked list by inorder traversal order.
 // Use postorder traversal to do the flattening job.
 #include <cstdio>
 using namespace std;

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

 void flatten(TreeNode *&root, TreeNode *&left_most, TreeNode *&right_most)
 {
     if (root == nullptr) {
         left_most = right_most = nullptr;
         return;
     }

     TreeNode *ll, *lr, *rl, *rr;
     if (root->left != nullptr) {
         flatten(root->left, ll, lr);
         root->left = lr;
         lr->right = root;
     } else {
         ll = lr = root;
     }

     if (root->right != nullptr) {
         flatten(root->right, rl, rr);
         root->right = rl;
         rl->left = root;
     } else {
         rl = rr = root;
     }

     left_most = ll;
     right_most = rr;
 }

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

     if (scanf("%d", &val) != ) {
         root = nullptr;
     } else if (val == ) {
         root = nullptr;
     } else {
         root = new TreeNode(val);
         constructBinaryTree(root->left);
         constructBinaryTree(root->right);
     }
 }

 void deleteList(TreeNode *&head)
 {
     TreeNode *ptr;

     while (head != nullptr) {
         ptr = head;
         head = head->right;
         delete ptr;
     }
 }

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

     while (true) {
         constructBinaryTree(root);
         if (root == nullptr) {
             break;
         }
         flatten(root, left_most, right_most);
         head = left_most;
         for (ptr = head; ptr != nullptr; ptr = ptr->right) {
             printf("%d ", ptr->val);
         }
         putchar('\n');
         deleteList(head);
     }

     return ;
 }