Binary Tree Traversals
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9221 Accepted Submission(s): 4167
Problem Description
A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2.
In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.
In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.
In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.
Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.
Input
The input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and inorder sequence. You can assume they are always correspond to a exclusive binary tree.
Output
For each test case print a single line specifying the corresponding postorder sequence.
Sample Input
9
1 2 4 7 3 5 8 9 6
4 7 2 1 8 5 9 3 6
Sample Output
7 4 2 8 9 5 6 3 1
前序遍历的第一个数据为树的根节点
而在中序遍历中，根结点的左边为左子树，右边为右子树
因此，先通过前序遍历确定树的根节点，再通过中序遍历确定左子树与右子树的范围，再将新得到的左右子树重复上述操作
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#include <cstdio> #include <string> #include <iostream> #define forr(a,b,c) for(int a=b;a<=c;a++) #define MAX(a,b) (a>b)?a:b; #define MIN(a,b) (a<b)?a:b; #define memsett(a) memset(a,sizeof(a),0); using namespace std; int pre[1001]; int in[1001]; bool boo; int post(int prel,int prer,int inl,int inr ) { int location; forr(i, inl, inr) if (in[i] == pre[prel]) { location = i; break; } if (location!=inl) post(prel + 1, prel + location - inl, inl, location - 1); if (location!=inr) post(prel + location - inl+1, prer, location + 1, inr); if (boo) printf(" "); printf("%d", pre[prel]); boo = true; return 0; } int main() { int n; while(scanf("%d", &n)==1){ boo = false; memsett(pre); memsett(in); forr(i, 1, n) scanf("%d", &pre[i]); forr(i, 1, n) scanf("%d", &in[i]); post(1,n,1,n); printf("\n"); } } |