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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.
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:
6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop
Sample Output:
3 4 2 6 5 1
代码如下：
#include#include #include #include #include using namespace std;const int maxn=35;int n,ci=0;struct tree{ int data; struct tree *left,*right;};tree* root;tree* traverse (){ if(ci==n*2) return NULL; char s[10]; scanf("%s",s); if(!strcmp(s,"Push")) { int x; scanf("%d",&x); tree* t=(tree*)malloc(sizeof(tree)); t->data=x; ci++; t->left=traverse(); ci++; t->right=traverse (); return t; } else return NULL;}void post (tree* t){ if(t==NULL) return; post(t->left); post(t->right); if(t==root) printf("%d\n",t->data); else printf("%d ",t->data);}int main(){ scanf("%d",&n); root=traverse(); post(root); return 0;}
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