PAT--1086. Tree Traversals Again

最新推荐文章于 2024-09-21 22:25:33 发布

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最新推荐文章于 2024-09-21 22:25:33 发布

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分类专栏： PAT 数据结构--二叉树文章标签： pat

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本文介绍了一种利用栈实现二叉树中序非递归遍历的方法，并通过给定的栈操作序列生成对应的二叉树。文章提供了一个示例输入输出，并附带完整的C++代码实现，用于计算该二叉树的后序遍历序列。

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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.
这里写图片描述

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

题解

入栈顺序就是先序序列，出栈顺序就是中序序列，构造后序序列。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 50;
int n;
int pre[maxn], in[maxn], post[maxn];
stack<int> sta;

void sol(int preIndex, int inIndex, int postIndex, int n){
    if(n == 0) return;
    if(n == 1) {
        post[postIndex] = pre[preIndex];
        return;
    }
    post[postIndex + n - 1] = pre[preIndex];
    int i = 0;
    for(; i < n && in[inIndex + i] != pre[preIndex]; ++i);
    int left = i, right = n - i - 1;
    sol(preIndex + 1, inIndex, postIndex, left);
    sol(preIndex + left + 1, inIndex + left + 1, postIndex + left, right);
}

int main(){
    cin >> n;
    string op; int index;
    int pre_k = 0, in_k = 0;
    for(int i = 0; i < 2 * n; ++i){
        cin >> op;
        if(op == "Push"){
            cin >> index;
            sta.push(index);
            pre[pre_k++] = index;
        }else{
            in[in_k++] = sta.top();
            sta.pop();
        }
    }

    sol(0, 0, 0, n);
    for(int i = 0; i < n; ++i){
        if(i) cout << " ";
        cout << post[i];
    }


    return 0;
}