编程题 03-树3 Tree Traversals Again【PAT】

编程练习题目集目录

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.

Figure 1

输入格式

  Each input file contains one test case. For each case, the first line contains a positive integer $N(\le 30)$ 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

题解

解题思路

  入栈是前序遍历，弹出是中序遍历，输出要求后续遍历。

完整代码

#include<iostream>
#include<string>
#include<vector>
#include<stack>using namespace std;// 递归函数：通过前序和中序遍历构造后序遍历
// 参数： preOrder：前序遍历数组 preL：前序遍历子数组起点 inOrder：中序遍历数组 nL：中序遍历子数组起点
//       postOrder：后序遍历结果数组（引用传递） postL：后序遍历子数组起点 n：当前子树节点数量
void getPostOrder(vector<int> preOrder, int preL, vector<int> inOrder, int inL, vector<int> &postOrder, int postL, int n)
{if (n == 0) return ;      // 递归终止条件：空树if (n == 1) {             // 叶子节点直接写入后序遍历postOrder[postL] = preOrder[preL];return ;}auto root = preOrder[preL];             // 根节点 = 前序遍历的第一个元素postOrder[postL + n - 1] = root;           // 后序遍历的根节点总是在最后int i = 0;                                 // 在中序数组中定位根节点位置while (i < n) {if (inOrder[inL + i] == root) break;++i;}int leftSize = i;                           // 左子树大小int rightSize = n - i - 1;                  // 右子树大小// 递归处理左子树// 前序数组：从preL+1开始，长度为leftSize 中序数组：从inL开始，长度为leftSize 后序数组：从postL开始，长度为leftSizegetPostOrder(preOrder, preL + 1, inOrder, inL, postOrder, postL, leftSize);// 递归处理右子树// 前序数组：从preL+leftSize+1开始，长度为rightSize 中序数组：从inL+leftSize+1开始，长度为rightSize 后序数组：从postL+leftSize开始，长度为rightSizegetPostOrder(preOrder, preL + leftSize + 1, inOrder, inL + leftSize + 1, postOrder, postL + leftSize, rightSize);
}// 模拟构造树的前序和中序遍历数组
vector<vector<int>> getOrder(int N)
{vector<int> preOrder(N, 0);         // 前序遍历数组vector<int> inOrder(N, 0);          // 中序遍历数组stack<int> stack;                        // 辅助栈int preIndex = 0, inIndex = 0;           // 索引指针for (int i = 0; i < 2*N; ++i) {          // 处理2N次操作string op; int val;cin >> op;if (op == "Push") {                  // 入栈操作（对应前序遍历）cin >> val;preOrder[preIndex++] = val;         // 记录前序遍历stack.push(val);                    // 元素入栈} else if (op == "Pop") {               // 出栈操作（对应中序遍历）inOrder[inIndex++] = stack.top();   // 记录中序遍历stack.pop();                        // 元素出栈}}return {preOrder, inOrder};         // 返回前序和中序遍历数组
}int main(void)
{int N;cin >> N;auto traversal = getOrder(N);       // 获取前序和中序遍历数组vector<int> preOrder = traversal[0];vector<int> inOrder = traversal[1];vector<int> postOrder(N, 0);                    // 初始化后序遍历数组getPostOrder(preOrder, 0, inOrder, 0, postOrder, 0, N);for (int i = 0; i < N; ++i) {                       // 输出后序遍历结果cout << postOrder[i];if (i < N-1) cout << " ";}cout << endl;return 0;
}