03-3. Tree Traversals Again (PAT) - 樹的遍歷問題

 

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

題意解析：入棧順序即為先序遍歷的順序，出棧順序即為中序遍歷的順序

解題思路：

    1.根據入棧出棧順序，建立先序遍歷數組與中序遍歷數組。

    2.取先序序列中的第一個元素，該元素為根結點

    3.根據根結點在中序序列中查找根結點的位置，從而得到該樹左子樹結點個數（L）與右子樹的結點個數（R）

    4.在后序序列數組中，第0到第L個元素為左子樹，第L+1到第L+R個元素為右子樹，最后一個元素為根結點

#include <iostream>
#include <string>
#include <stack>
using namespace std;

#define MAXSIZE 30

int preOrder[MAXSIZE];
int inOrder[MAXSIZE];
int postOrder[MAXSIZE];

void Solve(int preL, int inL, int postL, int n);

int main()
{
    for (int i = 0; i < MAXSIZE; i++)    //初始化數組
    {
        preOrder[i] = 0;
        inOrder[i] = 0;
        postOrder[i] = 0;
    }
    stack<int> inputStack;
    int nodeNum;
    cin >> nodeNum;
    int i, data;
    int preIndex = 0, inIndex = 0, postIndex = 0;
    string str;
    for (i = 0; i < 2 * nodeNum; i++)
    {
        cin >> str;
        if (str == "Push")
        {
            cin >> data;
            preOrder[preIndex++] = data;
            inputStack.push(data);
        }
        else if (str == "Pop")
        {
            inOrder[inIndex++] = inputStack.top();
            inputStack.pop();
        }
    }
    Solve(0, 0, 0, nodeNum);
    for (int i = 0; i < nodeNum; i++)
    {
        if ( i == nodeNum - 1 )
        {
            cout << postOrder[i] << endl;
        }
        else
        {
            cout << postOrder[i] << ' ';
        }
    }
}

void Solve(int preL, int inL, int postL, int n)
{
    if ( n == 0 )
    {
        return;
    }
    if ( n == 1 )
    {
        postOrder[postL] = preOrder[preL];
        return;
    }
    int root = preOrder[preL];
    postOrder[postL + n - 1] = root;
    int L, R;
    int i;
    for (i = 0; i < n; i++)
    {
        if ( inOrder[inL + i] == root )
        {
            break;
        }
    }
    L = i;    //左子樹結點個數
    R = n - L - 1;    //右子樹結點個數
    Solve(preL + 1, inL, postL, L);
    Solve(preL + L + 1, inL + L + 1, postL + L, R);
}