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:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

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提交代码

解题思路：

不管是左右子树还是整棵树，后序遍历的最后一个元素就是根节点，然后从中序遍历中找到根节点在中序遍历的位置，并且记录左子树元素的个数。

中序遍历中根节点以前的位左子树，以后的位右子树，然后递归遍建立左右子树，如果子树为空则返回0，l>r的情况就是子树为空的情况。在根据根节点输出层序遍历就可以了。

#include<iostream>  #include<cstdio>  #include<cstring>  #include<queue>  using namespace std;  #define N 35  int last[N],in[N];  struct node{      int left;      int right;  }tree[N];  int n,cnt;  int build(int l1,int r1,int l2,int r2){      int root;      if(l1 <= r1&&l2 <= r2){          root = last[r2];          int cnt  = 0;          int index;          for(index = l1;in[index] != root;index++,cnt++);//判定根节点下标和确定左子树的元素个数           tree[root].left = build(l1,index-1,l2,l2+cnt-1);//建立左子树           tree[root].right = build(index+1,r1,l2+cnt,r2-1);//建立右子树       }else{          return 0;      }      return root;  }  void print(int root){      queue<int>q;      q.push(root);//根节点入队       while(!q.empty()){          int temp = q.front();          cnt++;          printf("%d",temp);//输出结点           if(cnt != n){              printf(" ");          }else{              printf("\n");          }          q.pop();          if(tree[temp].left){              q.push(tree[temp].left);//左子树如队列           }          if(tree[temp].right){              q.push(tree[temp].right);//右子树入队列           }      }  }  int main(){  //  freopen("input.txt","r",stdin);      scanf("%d",&n);      for(int i = 1;i <= n;i++){          scanf("%d",&last[i]);      }      for(int i = 1;i <= n;i++){          scanf("%d",&in[i]);      }      int root = build(1,n,1,n);      print(root);      return 0;  }