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 = 0, inNo = 0, postNo = 0; 
    for(int i = 0; i < N * 2; 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(0, 0, 0, N);
    for(int i = 0; i < N; i++) {        //输出后序遍历序列  
        if(i == 0)                        //控制输出格式 
            printf("%d",postOrder[i]);
        else
            printf(" %d",postOrder[i]);
    }
    printf("
");
    return 0;
}

void postorderTraversal(int preNo, int inNo, int postNo, int N) 
{
    if(N == 0)
        return;
    if(N == 1) {
        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 -1] = root;     //后序遍历LRG最后一个为根
    for(int i = 0; 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 - 1;                        //右子树的结点数 
    postorderTraversal(preNo + 1, inNo, postNo, L);    //同理，将左子树看成新的树 
    postorderTraversal(preNo + L + 1, inNo + L + 1, postNo + L, R);//同理，右子树 
}