03-树3 Tree Traversals Again

二叉树及其遍历

push为前序遍历序列，pop为中序遍历序列。将题目转化为已知前序、中序，求后序。

前序GLR 中序LGR

前序第一个为G，在中序中找到G，左边为左子树L，右边为右子树R。

将左右子树看成新的树，同理。

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
Pop

Sample Output:

3 4 2 6 5 1

 #include <iostream>
 #include <cstdio>
 #include <stack>
 #include <string>
 using namespace std;

 #define MaxSize 30 

 #define OK 1
 #define ERROR 0

 int preOrder[MaxSize];
 int inOrder[MaxSize];
 int postOrder[MaxSize];

 void postorderTraversal(int preNo, int inNo, int postNo, int N);

 int main()
 {
     stack<int> stack;
     int N;        //树的结点数
     cin >> N;
     string str;
     int data;
     int preNo = , inNo = , postNo = ;
     for(int i = ; i < N * ; i++) {        //push + pop = N*2
         cin >> str;
         if(str == "Push") {            //push为前序序列
             cin >> data;
             preOrder[preNo++] = data;
             stack.push(data);
         }else{                        //pop出的是中序序列
             inOrder[inNo++] = stack.top();
             stack.pop();            //pop() 移除栈顶元素（不会返回栈顶元素的值）
         }
     }
     postorderTraversal(, , , N);
     for(int i = ; i < N; i++) {        //输出后序遍历序列
         if(i == )                        //控制输出格式
             printf("%d",postOrder[i]);
         else
             printf(" %d",postOrder[i]);
     }
     printf("\n");
     return ;
 }

 void postorderTraversal(int preNo, int inNo, int postNo, int N)
 {
     if(N == )
         return;
     if(N == ) {
         postOrder[postNo] = preOrder[preNo];
          return;
     }
     int L, R;
     int root = preOrder[preNo];            //先序遍历GLR第一个为根
     postOrder[postNo + N -] = root;     //后序遍历LRG最后一个为根
     for(int i = ; i < N; i++) {
         if(inOrder[inNo + i] == root) {    //找到中序的根 左边为左子树 右边为右子树
             L = i;                        //左子树的结点数
             break;
         }
     }
     R = N - L - ;                        //右子树的结点数
     postorderTraversal(preNo + , inNo, postNo, L);    //同理，将左子树看成新的树
     postorderTraversal(preNo + L + , inNo + L + , postNo + L, R);//同理，右子树
 }