1 题目

1086 Tree Traversals Again (25分)
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
PopSample Output:
3 4 2 6 5 1

2 解析

题意：根据栈的出栈、入栈序列，输出二叉树的后序序列
思路：
- 入栈序列为先序序列，出栈序列为中序序列
- 1 根据先序和中序序列建立二叉树，然后后序遍历输出二叉树

3 参考代码

#include <cstdio>
#include <iostream>
#include <stack>
#include <string>using std::cin;
using std::stack;
using std::string;const int MAXN = 35;
int in[MAXN];
int pre[MAXN];
int N;struct node
{int data;node* lchild;node* rchild;
};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] == root->data){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;
}int num = 0;
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);string str;int temp, num,inCount = 0,preCount = 0;stack<int> s;for (int i = 0; i < 2 * N; ++i){cin >> str;if(str == "Push"){scanf("%d", &num);s.push(num);pre[preCount++] = num;}else{temp = s.top();s.pop();in[inCount++] = temp;}}node* root = Create(0, N - 1, 0, N - 1);postorder(root);return 0;
}

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1086 Tree Traversals Again (25分)

1 题目

2 解析

3 参考代码

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