03-树3 Tree Traversals Again（25 point(s)） 【Tree】

03-树3 Tree Traversals Again（25 point(s)）

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

思路

题目给出 入栈的过程 要求求出 该树 的后序遍历结果

入栈过程 其实就是 前序遍历的结果

然后出栈过程 是中序遍历结果

所以实际上 题目的意思 就是 给出 前序遍历 和 中序遍历 求出 后序遍历

根据题给的样例

前序遍历 1 2 3 4 5 6
中序遍历 3 2 4 1 6 5

前序遍历 是 根 左 右
中序遍历 是 左 根 右
后序遍历 是 左 右 根

所以 我们可以认为 对于一棵树 来说 前序遍历的 第一个节点 就是它的根节点

然后 从 中序遍历 去找 直到 找到根节点，那么 根节点 前面的元素 就是左子树 右边的 元素 就是 右子树 然后 再去 前序遍历 找相应的元素区间 就是 对应的 左子树 区间 和 右子树 区间

其实下面的过程 就是 递归 重叠子问题的过程

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = 3.14159265358979323846264338327;
const double E = exp(1);
const double eps = 1e-30;

const int INF = 0x3f3f3f3f;
const int maxn = 1e3 + 5;
const int MOD = 1e9 + 7;

vector <int> pre, in, ans;

void post(int root, int start, int end)
{
    if (start >= end)
        return;
    int i = start;
    while (i < end && in[i] != pre[root])
        i++;
    int len = i - start;
    post(root + 1, start, i);
    post(root + len + 1, i + 1, end);
    ans.pb(pre[root]);
}

void print(vector <int> v)
{
    vector <int>::iterator it;
    for (it = v.begin(); it != v.end(); it++)
    {
        if (it != v.begin())
            printf(" ");
        printf("%d", (*it));
    }
    printf("\n");
}

int main()
{
    stack <int> st;
    int n;
    scanf("%d", &n);
    int m = n << 1;
    string s;
    int num;
    for (int i = 0; i < m; i++)
    {
        cin >> s;
        if (s == "Push")
        {
            scanf("%d", &num);
            st.push(num);
            pre.pb(num);
        }
        else
        {
            in.pb(st.top());
            st.pop();
        }
    }
    post(0, 0, n);
    print(ans);
}