1086. Tree Traversals Again (25)

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本文介绍了一种通过栈操作实现非递归中序遍历二叉树的方法，并根据给定的Push和Pop操作序列生成对应的二叉树后序遍历序列。文章提供了完整的C++代码实现，帮助读者理解算法细节。

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


题目意思：中序可以由栈完成。现在给出Push和Pop的操作，求出二叉树的后序。
思路：用一个数组记录节点，Pop的记录为0，Push就记录原本值，然后dfs生成树，然后递归生成后序顺序输出。给出N（30）个节点，如果全部是右子树
用数组得分配特别的内存。用个map记录。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
const int maxn = 70;
int n, tot;
int node[maxn], post[maxn];
map<int, int> tree;
void dfs(int idx)
{
    if(node[tot] == 0){
        ++tot;
        return;
    }
    tree[idx] = node[tot++];
    dfs(2 * idx + 1);
    dfs(2 * idx + 2);
}
void postOrder(int idx)
{
    if(tree.count(idx) == 0)
        return;
    postOrder(2 * idx + 1);
    postOrder(2 * idx + 2);
    post[tot++] = tree[idx]; 
}
int main()
{
    scanf("%d", &n);
    char str[5];
    int key;
    for(int i = 0; i < 2 * n; i++){
        scanf("%s", str);
        if(strcmp(str, "Push") == 0){
            scanf("%d", &key);
            node[i] = key;
        }
    }
//    for(int i = 0; i < 2 * n; i++)
//        printf("%d\n", node[i]);
    dfs(0);
    tot = 0;
    postOrder(0);
    for(int i = 0; i < n; i++){
        if(i)
            printf(" ");
        printf("%d", post[i]);
    }
    return 0;
}