PAT 1086
An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.
Figure 1Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.
Output Specification:
For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input: 6 Push 1 Push 2 Push 3 Pop Pop Push 4 Pop Pop Push 5 Push 6 Pop Pop Sample Output: 3 4 2 6 5 1 题意：输入按照栈的形式，输入先序和中序，输出后序 #include <cstdio> #include <iostream> #include <cstring> #include <stack> #include <algorithm> using namespace std; const int maxn=50; struct node{ int data; node* ld; node* rd; }; int pre[maxn],in[maxn],post[maxn]; int n; node* create(int prel,int prer,int inl,int inr) //由先序中序构建二叉树 { if(prel>prer) return NULL; node* root=new node; root->data =pre[prel]; int k; for(k=inl;k<=inr;k++) { if(in[k]==pre[prel]) break; } int numleft=k-inl; root->ld =create(prel+1,prel+numleft,inl,k-1); root->rd =create(prel+numleft+1,prer,k+1,inr); return root; } int num; void postorder(node* root) //后序输出 { if(root==NULL) return; postorder(root->ld); postorder(root->rd); printf("%d",root->data); num++; if(num<n) printf(" "); } int main() { scanf("%d",&n); char str[5]; stack<int> st; int x,prei=0,ini=0; for(int i=0;i<2*n;i++) //输入处理，把栈转化为先序数组和中序数组 { scanf("%s",str); if(strcmp(str,"Push")==0) { scanf("%d",&x); pre[prei++]=x; st.push(x); } else { in[ini++]=st.top(); st.pop() ; } } node* root=create(0,n-1,0,n-1); //构建二叉树 postorder(root); //后序输出 return 0; }