1086. Tree Traversals Again (25)
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.
![](http://nos.patest.cn/bs_n9mde9jcnyj.jpg)
Figure 1
Input 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
#include<cstdio> #include<stack> #include<cstring> using namespace std; const int maxn = 40; struct node{ int data; node* lchild; node* rchild; }; int pre[maxn],in[maxn],post[maxn]; int n; node* create(int preL,int preR,int inL,int inR){ 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; //输出数值不对要检查判断数值问题 } int numleft = k - inL; root->lchild = create(preL+1,preL+numleft,inL,k-1); root->rchild = create(preL+numleft+1,preR,k+1,inR); return root; } int num = 0; void postorder(node* root){ if(root == NULL) return; postorder(root->lchild); postorder(root->rchild); printf("%d",root->data); num++; if(num < n)printf(" "); } int main(){ scanf("%d",&n); stack<int> st; int x,preIndex = 0,inIndex = 0; char str[10]; for(int i = 0; i < 2*n; i++){ scanf("%s",str); if(strcmp(str,"Push") == 0){ scanf("%d",&x); st.push(x); pre[preIndex++] = x; }else{ in[inIndex++] = st.top(); st.pop(); } } node* root = create(0,n-1,0,n-1); postorder(root); return 0; }