03-树3 Tree Traversals Again (25 分)
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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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 = ;
struct Node{
int data;
Node* lchild;
Node* rchild;
};
int n,pre[maxn],in[maxn],num = ; Node* createTree(int preL,int preR,int inL,int inR){
if(preL > preR) return NULL;
Node* root = new Node;
root -> data = pre[preL];
//printf("%d\n",root->data);
int k;
for(k = inL; k <= inR; k++){
if(in[k] == pre[preL]) break;
}
int numLeft = k - inL;
root->lchild = createTree(preL+,preL+numLeft,inL,k-);
root->rchild = createTree(preL+numLeft+,preR,k+,inR);
return root;
} 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(){
int x,k1=,k2=;
scanf("%d",&n);
stack<int> st;
char str[];
for(int i = ; i < *n; i++){
scanf("%s",str);
if(strcmp(str,"Push") == ){
scanf("%d",&x);
st.push(x);
pre[k1++] = x;
}else{
in[k2++] = st.top();
st.pop();
}
}
// printf("1\n");
Node* root = createTree(,n-,,n-);
// printf("2\n");
postOrder(root);
return ;
}
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