PAT甲级.1086. Tree Traversals Again (25)

题目

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.

输入格式

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.

输出格式

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.

输入样例

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

输出样例

3 4 2 6 5 1

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PAT链接

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思路

1.Push的次序为1、2、3、4、5、6，即先序遍历序列为123456
2.Pop序列为3，2，4，1，6，5 即中序遍历序列为324165
Push的序列为先序序列，Pop的序列为后序序列，具体由stack实现
另外关键是由前序序列和中序序列重建二叉树

代码

/**
* @tag     PAT_A_1086
* @authors R11happy ([email protected])
* @date    2016-10-22 4:56-6:10
* @version 1.0
* @Language C++
* @Ranking  670/1674
* @function null
*/

#include 
#include 
#include 
#include 
#include 
using namespace std;


struct node
{
    int data;
    node *lchild;
    node *rchild;
};

int n;
int num;
int pre[40], in[40];

node *create(int preL, int preR, int inL, int inR)
{
    if (preL > preR) return NULL;
    node *root = new node;
    root->data = pre[preL];

    int k;
    for (k = inL; k <= inR; k++)
    {
        if (in[k] == pre[preL]) break;
    }

    int numLeft = k - inL;
    root->lchild = create(preL + 1, preL + numLeft, inL, k - 1);
    root->rchild = create(preL + numLeft + 1, preR, k + 1, inR);
    return root;    //不要忘记返回root
}

void postOrder(node *root)
{
    if (root == NULL)    return;
    postOrder(root->lchild);
    postOrder(root->rchild);
    printf("%d", root->data);
    num++;
    if (num < n) printf(" ");
}

int main(int argc, char const *argv[])
{
    scanf("%d", &n);
    char str[5];
    stack<int> st;
    int x, preIndex = 0, inIndex = 0;
    for (int i = 0; i2; i++)
    {
        scanf("%s", str);
        if (strcmp(str, "Push") == 0)
        {
            scanf("%d", &x);
            pre[preIndex++] = x;
            st.push(x);
        }
        else
        {
            in[inIndex++] = st.top();
            st.pop();
        }
    }

    node *root = create(0, n - 1, 0, n - 1);
    postOrder(root);
    return 0;
}

/*
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
 */

收获

1.入栈出栈操作

        if (strcmp(str, "Push") == 0)
        {
            scanf("%d", &x);
            pre[preIndex++] = x;
            st.push(x);
        }
        else
        {
            in[inIndex++] = st.top();
            st.pop();
        }

2.前序序列+中序序列重建二叉树

node *create(int preL, int preR, int inL, int inR)
{
    if (preL > preR) return NULL;
    node *root = new node;
    root->data = pre[preL];

    int k;
    for (k = inL; k <= inR; k++)
    {
        if (in[k] == pre[preL]) break;
    }

    int numLeft = k - inL;
    root->lchild = create(preL + 1, preL + numLeft, inL, k - 1);
    root->rchild = create(preL + numLeft + 1, preR, k + 1, inR);
    return root;    //不要忘记返回root
}