首页 > 代码库 > 《Cracking the Coding Interview》——第17章：普通题——题目13

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

2024-07-02 00:19:36   229人阅读

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 }

 


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