A1086. Tree Traversals Again
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
#include<cstdio>
#include<iostream>
#include<stack>
#include<string.h>
using namespace std;
typedef struct NODE{
NODE* left, *right;
int data;
}node;
stack<int> stk;
int pre[], in[], 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 i;
for(i = inL; i <= inR; i++)
if(in[i] == root->data)
break;
int Lnum = i - inL;
root->left = create(preL + , preL + Lnum, inL, i - );
root->right = create(preL + Lnum + , preR, i + , inR);
return root;
}
void post(node *tree, int &cnt){
if(tree == NULL)
return;
post(tree->left, cnt);
post(tree->right, cnt);
if(cnt == N - )
printf("%d", tree->data);
else{
printf("%d ", tree->data);
cnt++;
}
} int main(){
int num, indexPre = , indexIn = ;
char str[];
scanf("%d", &N);
for(int i = ; i < *N; i++){
scanf("%s", str);
if(strcmp(str, "Push") == ){
scanf("%d ", &num);
stk.push(num);
pre[indexPre++] = num;
}else{
num = stk.top();
stk.pop();
in[indexIn++] = num;
}
}
node *tree = create(, N - , , N - );
int cnt = ;
post(tree, cnt);
cin >> N;
return ;
}
总结:
1、中序遍历的非递归实现:不断将非空的左孩子入栈,当左边为空时,弹出一个栈顶元素访问之,并将指针移至他的右子树,继续进行开始时的操作。 在这个过程中,push的特点是不断把遇到的新节点push入栈,因此push的过程就是先序遍历的过程。而pop自然是中序的过程。因此,中序与先序的非递归实现的区别就在于访问节点的时机不同,先序在push时,中序在pop时。
2、因此可以根据题目同时建立一个栈,同步做push和pop操作,先得到先序与中序遍历序列,再按照套路建树。
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