PTA 03-树3 Tree Traversals Again

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(≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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

题目大意：

  根据所给的对树的操作，给出树的后序遍历

题目分析：

  push的顺序即是树的前序遍历，然后再结合pop可以得出该树的中序遍历，利用push和pop的操作顺序，来重新构建树，再利用递归对树进行后续遍历。

  当push一个节点时，若该push操作的前一个操作是push操作，则该节点作为parent节点的左儿子；若该push操作的前一个操作为pop操作，则该节点作为parent节点的右儿子。

代码：

#include<stdio.h>
#include<stdlib.h>
#define ElementType int

#define STR_LEN 5
typedef struct TreeNode *Tree;
struct TreeNode{
  ElementType Element;
  Tree Left;
  Tree Right;
};

struct Stack{
  Tree location;
}stack[100];

int top=0;
ElementType values[20];
int num=0;

void Push(Tree temp)
{
  stack[top].location=temp;
  top++;
}
Tree Pop()
{
  if(top)
   {
       Tree node=stack[--top].location;
       return node;
   }
   else
    return NULL;
}

Tree BuildTree()
{
    char str[STR_LEN];
    int N,i,num;
    int poped=0,findRoot=0;
    Tree parent=NULL,Root=NULL;
    scanf("%d",&N);
    for(i=0;i<2*N;i++)
    {
        scanf("%s",&str);
        if(strcmp(str,"Push")==0)
        {
            scanf("%d",&num);
            Tree node=(Tree)malloc(sizeof(struct TreeNode));
            node->Element=num;
            node->Left=NULL;
            node->Right=NULL;
            //if(!parent&&!poped)
            if(parent&&!poped) //父节点不为空，但前一个操作为push，则新输入节点作为父节点的左孩子
            {
                parent->Left=node;
                parent=node;
            }
            else if(parent&&poped) //前一个操作为pop，则新输入的节点作为父节点的右节点
            {
                parent->Right=node;
                parent=node;
            }
            else //父节点为空，说明还未添加节点到树中
            {
                 Root=node;
                 parent=node;
            }
            Push(parent);
            poped=0;
        }
        else
        {
            parent=Pop();
            poped=1;
        }
    }
    return Root;
}

void PostOrder(Tree T)
{
    Tree tn = T;
    if(tn)
    {
        PostOrder(tn->Left);
        PostOrder(tn->Right);
        values[num++] = tn->Element; //将后序遍历出的节点值存入数组便于格式化打印
    }
}

int main()
{
   Tree T=BuildTree();
   Tree middle;
   PostOrder(T);
   int i;
   for (i = 0; i < num-1; i++)
      printf("%d ", values[i]);
   printf("%d\n", values[num-1]);
    return 0;


            }