1020. Tree Traversals

1020. Tree Traversals (25)

时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2
  1 #include <iostream>
2 #include <fstream>
3 #include <vector>
4 #include <string>
5 #include <algorithm>
6 #include <map>
7 #include <stack>
8 #include <cmath>
9 #include <queue>
10 #include <set>
11
12
13 using namespace std;
14
15 class Node
16 {
17 public:
18 Node* left;
19 Node* right;
20 int value;
21 };
22
23 Node* getNode( int inorder[30] , int postorder[30] , int i_base , int i_end, int p_base , int p_end )
24 {
25 if( i_end > i_base && p_end > p_base )
26 {
27 Node *center = new Node();
28
29 center->value = postorder[ p_end ];
30
31 int i_left_end = i_base;
32
33 for( ; i_left_end <= i_end ; ++i_left_end )
34 {
35 if( inorder[i_left_end] == center->value )
36 {
37 break;
38 }
39 }
40
41 --i_left_end;
42
43 if( i_left_end >= i_base )
44 {
45 int p_left_end = i_left_end - i_base + p_base;
46
47 center->left = getNode( inorder , postorder, i_base , i_left_end , p_base , p_left_end );
48 }
49 else
50 {
51 center->left = NULL;
52 }
53
54 int i_right_base = i_left_end + 2;
55
56 if( i_right_base <= i_end )
57 {
58 int p_right_base = p_end - 1 - ( i_end - i_right_base );
59
60 center->right = getNode( inorder , postorder, i_right_base , i_end , p_right_base , p_end - 1 );
61 }
62 else
63 {
64 center->right = NULL;
65 }
66
67 return center;
68 }
69 else if( i_end == i_base && p_end == p_base )
70 {
71 Node *center = new Node();
72
73 center->value = postorder[ p_end ];
74
75 center->left = NULL;
76 center->right = NULL;
77 return center;
78 }
79 else
80 {
81 return NULL;
82 }
83
84
85 }
86
87
88 int main()
89 {
90
91
92 int N;
93
94 int p_order[30];
95 int i_order[30];
96
97
98 while( cin >> N )
99 {
100 for( int i = 0 ; i < N ; ++i )
101 {
102 cin >>p_order[i];
103 }
104
105 for( int i = 0 ; i < N ; ++i )
106 {
107 cin >>i_order[i];
108 }
109
110 Node* tree = getNode( i_order , p_order , 0 , N -1 , 0 , N - 1 );
111
112 queue<Node*> queue;
113
114 queue.push(tree);
115
116 bool first = true;
117 while( !queue.empty() )
118 {
119 if(!first)
120 {
121 cout << " ";
122 }
123 else
124 {
125 first = false;
126 }
127
128 cout << queue.front()->value;
129
130 if( queue.front()->left != NULL )
131 {
132 queue.push(queue.front()->left);
133 }
134
135 if(queue.front()->right != NULL)
136 {
137 queue.push(queue.front()->right);
138 }
139
140 queue.pop();
141 }
142 cout << endl;
143
144 }
145
146 return 0;
147 }


原文地址:https://www.cnblogs.com/kking/p/2331815.html