1020. Tree Traversals (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1020

题目：

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 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

分析：

题目要求的是依据树的后序遍历和中序遍历构造树，再层序遍历输出。这就考察我们三点：

1）树的结构体会不会，新节点的生成等等。

2）两种遍历构造树会不会，用递归函数写

3）层序遍历会不会，用队列辅助（层序、广搜用队列；深搜用递归、栈）

AC代码：

#include<stdio.h>
#include<queue>
using namespace std;
struct Node{//节点结构体
 Node *lchild;
 Node*rchild;
 int c;
}Tree[32];
int loc;
int str1[32], str2[32];//分别存储后序遍历和中序遍历的字符串
Node *create(){
 Tree[loc].lchild = Tree[loc].rchild = NULL;
 return &Tree[loc++];
}
queue<Node *>Q;//用于层序遍历，用队列
Node *build(int s1, int e1, int s2, int e2){//利用递归来构建数
 Node *ret = create();
 ret->c = str1[e1];
 int rootIdx;
 for (int i = s2; i <= e2; i++){//由兴许遍历找到根节点，然后找到其在中序的位置
  if (str2[i] == str1[e1]){
   rootIdx = i;
   break;
  }
 }
 if (rootIdx != s2){//然后进行左右子树的分隔
  ret->lchild = build(s1, s1 + rootIdx - s2 - 1, s2, rootIdx - 1);
 }
 if (rootIdx != e2){
  ret->rchild = build(s1 + rootIdx - s2, e1 - 1, rootIdx + 1, e2);
 }
 return ret;
}
int main(void){
 //freopen("F://Temp/input.txt", "r", stdin);
 int n;
 while (scanf("%d", &n) != EOF){
  for (int i = 0; i < n; i++){
   scanf("%d", &str1[i]);
  }
  for (int i = 0; i < n; i++){
   scanf("%d", &str2[i]);
  }
  loc = 0;
  Node *T = build(0, n - 1, 0, n - 1);
  Q.push(T);
  while (!Q.empty()){//用队列来层序遍历输出。把对头元素取出输出，并让左右节点（假设有）入队
   Node *tmp = Q.front();
   if (tmp -> lchild != NULL){
    Q.push(tmp->lchild);
   }
   if (tmp -> rchild != NULL){
    Q.push(tmp -> rchild);
   }
   Q.pop();
   if (Q.empty()){
    printf("%d
", tmp -> c);
   }
   else printf("%d ", tmp -> c);
  }
 }
 return 0;
}

截图：

——Apie陈小旭