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

#include <stdio.h>
#include <string.h> 

void ReadTree(int Pre[], int In[], int N);
void GetPost(int Pre[], int In[], int Post[], int preL,
                   int inL, int postL, int N);

int main(){
	int N;
	scanf("%d",&N);
	int flag = 0,Pre[30]={0},In[30]={0},Post[30]={0};
	ReadTree(Pre, In, N); 
	GetPost(Pre, In, Post, 0, 0, 0 ,N);
	for (int i = 0; i < N; ++i)
	{
		if(!flag) 
			{printf("%d",Post[i]);
			 flag++;
			}
		else printf(" %d",Post[i]);
	}
	return 0;
}
void ReadTree(int Pre[], int In[], int N){
	int tail1=0,tail2=0, flag=0, A[30]={0};
	int Stack[30]={0}; 
	char op[4];
	for(int i = 0; i< 2*N ; i++){
		scanf("%s",&op);
		if( ! strcmp(op,"Push")) 
			{ scanf("%d",&Stack[flag]);               //构造前序遍历数组
			  Pre[tail1++] = Stack[flag];
			  flag++;
			}
		else  In[tail2++] = Stack[--flag];
							                         //构造中序遍历数组
}
}


void GetPost(int Pre[], int In[], int Post[], int preL,
                   int inL, int postL, int N)
{
    if (N == 0) return ;
    if (N == 1) {
        Post[postL] = Pre[preL];
        return ;
    }
    int root = Pre[preL];
    Post[postL + N - 1] = root;
    //在中序遍历数组上找出root的位置
    int i = 0;
    while (i < N) {
        if (In[inL + i] == root) break;
        ++i;
    }
    // 计算出root节点左右子树节点的个数
    int L = i, R = N - i - 1;
    // 递归的进行求解
    GetPost(Pre, In , Post, preL + 1, inL, postL, L);
    GetPost(Pre, In, Post, preL + L + 1, inL + L + 1, postL + L, R);
}

总结：

1.入栈顺序即为前序遍历，出栈顺序为中序遍历

2.  先读入并存储前序遍历数组和中序遍历数组，然后求后序遍历

3.   前序第一个值即为Root，也是后序数组最后一个值，同时Root将中序数组分割为左子树和右子树。 然后递归求解左子树和右子树。

对于Sample的求解结果为： 

 
刚开始时，前序遍历的第一个元素1肯定是该二叉树的根节点，因此为postOrder的最后一个元素，在中序遍历中1前面的元素肯定为该颗二叉树的左子树在1右边的元素肯定为该二叉树的右子树，因此，递归的求解该二叉树。 

 

参考资料：https://blog.youkuaiyun.com/shinanhualiu/article/details/49279051