03-树3 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 NN (\le 30≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to NN). Then 2N2N 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

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AC代码:

#include <iostream>using namespace std;struct Node{    int tag;   //第几次进栈     int num;};/*先序遍历对应进栈顺序，中序遍历对应出栈顺序；    后序遍历与中序遍历不同的是节点出栈后要马上再入栈（tag做第二次入栈标记），等右儿子遍历完后再出栈;    具体实现上，每次中序遍历的pop时，如果栈顶是标记过的节点（tag=2），循环弹出；如果没有标记过（tag=1），做标记，即弹出再压栈）    栈顶tag=2的节点对应中序遍历中已弹出的节点；循环弹出后碰到的第一个tag=1的节点才对应中序遍历当前pop的节点  */int main(){    int N;    cin>>N;    struct Node stack[30];    int flag = 0;    int size = 0;               //栈元素大小，指向栈顶的下一个位置      for ( int i = 0 ; i < 2*N ; i++){        string str;        cin>>str;        if(str[1] == 'u'){            cin>>stack[size].num;   //入栈              stack[size].tag = 1;    //标记第一次入栈              size++;        }else{            //循环弹出栈顶tag=2的节点             while(size > 0 && stack[size-1].tag == 2){                if( flag){                    cout<<" ";                }                flag = 1;                cout<<stack[--size].num;            }            //将中序遍历中应该要弹出的节点弹出再压栈，做标记即可              if ( size>0){                stack[size-1].tag = 2;            }        }     }    while(size){        if ( flag ){            cout<<" ";        }        flag = 1;        cout<<stack[--size].num;    }    return 0;}