【PAT甲级 - C++题解】1086 Tree Traversals Again

1086 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

题意

通过使用栈可以以非递归方式实现二叉树的中序遍历。

例如，假设遍历一个如下图所示的 6 66 节点的二叉树（节点编号从 1 11 到 6 66）。

则堆栈操作为：push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop() 。

我们可以从此操作序列中生成唯一的二叉树。

你的任务是给出这棵树的后序遍历。

思路

可以从用栈实现二叉树的中序遍历的思路中找到规律：

如果上一次操作是 Push ，则当前 Push 的结点是其父结点的左孩子，这里的父结点是上次 Push 的结点。例如，push(1); push(2); ，则结点 2 是上次 Push 的结点 1 的左孩子。

如果上一次操作是 Pop ，则当前 Push 的结点是其父结点的右孩子，这里的父节点是上次 Pop 的结点。例如，push(1); push(2); push(3); pop(); pop(); push(4); ，结点 4 是上次 Pop 的结点 2 的右孩子。

如果上一次没有操作，则当前 Push 的结点就是根结点。

代码

#include<bits/stdc++.h>
using namespace std;
const int N = 40;
int n;
int l[N], r[N]; //用哈希表来存储每个结点左孩子的下标
void dfs(int u, int root)
{
    if (!u)  return;
    dfs(l[u], root);
    dfs(r[u], root);
    cout << u;
    if (u != root) cout << " "; //最后一个结点输出末尾不能加空格
}
int main()
{
    int last = 0, type, root;
    stack<int> stk;
    cin >> n;
    for (int i = 0; i < n * 2; i++)
    {
        string op;
        cin >> op;
        if (op == "Push")
        {
            int x;
            cin >> x;
            if (!last)   root = x; //第一个结点为根结点
            else    //设置父结点的左右孩子
            {
                if (type == 0) l[last] = x;  //上个操作是Push
                else    r[last] = x;  //上个操作是Pop
            }
            last = x;
            type = 0; //表示Push
            stk.push(x);
        }
        else
        {
            last = stk.top();
            stk.pop();
            type = 1; //表示Pop
        }
    }
    dfs(root, root); //后序遍历
    return 0;
}