PTA03-树3 Tree Traversals Again

先写下题目吧。

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 $N$ ($\le 30$) 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:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1

这道题在之前陈越老师还没出讲解的时候做了一遍，大致思路是根据出给的前序和中序得到两个数列，再根据遍历算法生成二叉树。这里我只贴出根据前序和中序数列建二叉树的过程。

BinTree BuildTree(int preorder[],int pre_s,int pre_e,int inorder[],int in_s,int in_e) //preorder和inorder分别为前序和中序的数列，pre_s，in_s分别是数列起点

//pre_e和in_e分别为数列最后元素所在位置{int i;BinTree p;if(in_s>in_e)return NULL;for(i=in_s;i<in_e&&inorder[i]!=preorder[pre_s];i++);p=(BinTree)malloc(sizeof(struct TreeNode));p->Data=preorder[pre_s];p->Left=BuildTree(preorder,pre_s+1,pre_s+i-in_s,inorder,in_s,i-1);p->Right=BuildTree(preorder,pre_s+i-in_s+1,pre_e,inorder,i+1,in_e);return p;}

主要的思想是分而治之。

今天听了这道题讲解，多了一个新思路，不建树，根据前序遍历的第一个元素即是后序遍历的最后一个元素，分而治之，下面给出我写的不用建树，直接得出后序遍历数列的程序。

void posttraversal2(int preorder[],int pre_s,int pre_e,int inorder[],int in_s,int in_e){int i;if(in_s>in_e)return ;postorder[count_3]=preorder[pre_s];count_3--;//初始值为前序遍历的最后一个元素的下标for(i=in_s;i<in_e&&inorder[i]!=preorder[pre_s];i++);posttraversal2(preorder,pre_s+i-in_s+1,pre_e,inorder,i+1,in_e);posttraversal2(preorder,pre_s+1,pre_s+i-in_s,inorder,in_s,i-1);return ;}

其实这部分程序我直接根据上面建树的部分改写的，主要一点是得先遍历右子树，再遍历左子树。我测了一下，答案应该是一样的