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[31], in[31], 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 + 1, preL + Lnum, inL, i - 1);
    root->right = create(preL + Lnum + 1, preR, i + 1, inR);
    return root;
}
void post(node *tree, int &cnt){
    if(tree == NULL)
        return;
    post(tree->left, cnt);
    post(tree->right, cnt);
    if(cnt == N - 1)
        printf("%d", tree->data);
    else{
        printf("%d ", tree->data);
        cnt++;
    }
}

int main(){
    int num, indexPre = 0, indexIn = 0;
    char str[5];
    scanf("%d", &N);
    for(int i = 0; i < 2*N; i++){
        scanf("%s", str);
        if(strcmp(str, "Push") == 0){
            scanf("%d ", &num);
            stk.push(num);
            pre[indexPre++] = num;
        }else{
            num = stk.top();
            stk.pop();
            in[indexIn++] = num;
        }
    }
    node *tree = create(0, N - 1, 0, N - 1);
    int cnt = 0;
    post(tree, cnt);
    cin >> N;
    return 0;
}

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。