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：
1 // 17.13 Flatten a binary search tree into a doubly linked list by inorder traversal order.
2 // Use postorder traversal to do the flattening job.
3 #include <cstdio>
4 using namespace std;
5
6 struct TreeNode {
7 int val;
8 TreeNode *left;
9 TreeNode *right;
10 TreeNode(int _val = 0): val(_val), left(nullptr), right(nullptr) {};
11 };
12
13 void flatten(TreeNode *&root, TreeNode *&left_most, TreeNode *&right_most)
14 {
15 if (root == nullptr) {
16 left_most = right_most = nullptr;
17 return;
18 }
19
20 TreeNode *ll, *lr, *rl, *rr;
21 if (root->left != nullptr) {
22 flatten(root->left, ll, lr);
23 root->left = lr;
24 lr->right = root;
25 } else {
26 ll = lr = root;
27 }
28
29 if (root->right != nullptr) {
30 flatten(root->right, rl, rr);
31 root->right = rl;
32 rl->left = root;
33 } else {
34 rl = rr = root;
35 }
36
37 left_most = ll;
38 right_most = rr;
39 }
40
41 void constructBinaryTree(TreeNode *&root)
42 {
43 int val;
44
45 if (scanf("%d", &val) != 1) {
46 root = nullptr;
47 } else if (val == 0) {
48 root = nullptr;
49 } else {
50 root = new TreeNode(val);
51 constructBinaryTree(root->left);
52 constructBinaryTree(root->right);
53 }
54 }
55
56 void deleteList(TreeNode *&head)
57 {
58 TreeNode *ptr;
59
60 while (head != nullptr) {
61 ptr = head;
62 head = head->right;
63 delete ptr;
64 }
65 }
66
67 int main()
68 {
69 TreeNode *root;
70 TreeNode *left_most, *right_most;
71 TreeNode *head;
72 TreeNode *ptr;
73
74 while (true) {
75 constructBinaryTree(root);
76 if (root == nullptr) {
77 break;
78 }
79 flatten(root, left_most, right_most);
80 head = left_most;
81 for (ptr = head; ptr != nullptr; ptr = ptr->right) {
82 printf("%d ", ptr->val);
83 }
84 putchar(‘\n‘);
85 deleteList(head);
86 }
87
88 return 0;
89 }
《Cracking the Coding Interview》——第17章：普通题——题目13,码迷,mamicode.com
时间： 2024-12-19 08:42:46