PTA 5-5 Tree Traversals Again (25) - 树 - 二叉树及其遍历

题目：http://pta.patest.cn/pta/test/16/exam/4/question/667

PTA - Data Structures and Algorithms (English) - 5-5

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.

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 //即后序遍历树的结果

解法转自：http://www.cnblogs.com/clevercong/p/4177802.html

题目分析：

动态的“添加结点”并遍历过程，不需要建树。

借助“栈”进行树的后续遍历；( times：可由此栈操作判断中当前结点有左子树还是右子树中 )

每次PUSH，times = 1；// PUSH：可看作访问左孩子

每次POP检查栈顶记录的times：

- 如果是1，弹出来变成2压回栈；// POP：可看作由左孩子转到访问右孩子

- 如果是2，则弹出，放入存放结果的vector中，重复这一过程，直到栈顶times为1。// POP：可看作访问根节点

所有PUSH与POP操作执行完毕时，输出vector内的数据和stack中的数据。 注意要处理最后的空格。

注：此方法可推广到n叉树，只需改变times的值即可。

            ……

代码：

#include <iostream> #include <stack> #include <vector> using namespace std; typedef struct node { int data; int times; node(int d, int t) :data(d), times(t) {}; } Node; int main() { int n; cin >> n; string cmd; //!PUSH or POP int x; stack<Node> sta; //!栈：用来实现后序遍历 vector<int> vec; //!存放结果 for(int i=0; i<2*n; i++) { cin >> cmd; if(cmd == "Push") { cin >> x; sta.push(Node(x, 1)); } if(cmd == "Pop") { Node node = sta.top(); sta.pop(); if(node.times == 1) //!times==1：弹出来变成2压回栈 { node.times = 2; sta.push(node); } else if(node.times == 2) //!times==2：弹出，放入存放结果的vector中 { vec.push_back(node.data); while(sta.top().times == 2) //!重复这一过程，直到栈顶的times==1 { vec.push_back(sta.top().data); sta.pop(); } if (sta.size() != 0) //!操作结束后栈还不为空，使栈顶的times=2 { sta.top().times=2; } } } } for(unsigned int i=0; i<vec.size(); i++) { cout << vec[i]<< " "; } while(sta.size() != 0) //!PUSH/POP结束后栈中还有元素，直接pop()即可 { cout << sta.top().data; sta.pop(); if(sta.size() != 0) cout << " "; } return 0; }

转载于:https://www.cnblogs.com/claremore/p/4806304.html