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 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

//typedef long long LL; const int maxn = 50; struct node{ int data; node* lchild; node* rchild; }; int pre[maxn], in[maxn], post[maxn];//先序，中序以及后序 int n;//结点个数

//当前二叉树的先序序列区间为[preL, preR],中序序列区间为[inL, inR] //create函数返回根节点 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; } 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;//已输出的结点个数 void postorder(node* root){ //后序遍历 if (root == NULL) { return; } postorder(root->lchild); postorder(root->rchild); printf("%d", root->data); num++; if(num < n) printf(" "); } int main(){ scanf("%d", &n); char str[5]; stack<int> st; int x, preIndex = 0, inIndex = 0;//入栈元素、先序序列位置及中序序列位置 for (int i = 0; i < 2 * n; i++) { scanf("%s", str); if (strcmp(str, "Push") == 0) {//入栈 scanf("%d", &x); pre[preIndex++] = x; st.push(x); }else{ in[inIndex++] = st.top(); st.pop(); } } node* root = create(0, n-1, 0, n-1); postorder(root); return 0; }