《数据结构》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 (≤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

分析

题目大意就是中序非递归遍历需要用到一个栈，给定针对这个栈的操作，就可以唯一确定一个树，我们的任务就是后序遍历这棵树
回忆一下如何利用栈进行中序遍历，首先根入栈，其左结点循环入栈，如果此时栈不空，出栈栈顶结点并输出，再让栈顶结点等于其右子结点，直到整个栈为空且结点为空
所以对于 push 操作要做的事有两种可能：

1. 入栈当前结点的左儿子结点
2. 入栈当前结点的右儿子结点

好像说了废话，想想非递归遍历的操作，如果左儿子结点一直都有，那么入栈的一直是左儿子结点，只有当左儿子结点已经没有了，才"勉为其难"入栈右儿子结点。
而对于 pop 操作要做的事也有两种可能：

1. 当前结点的左儿子结点为空时出栈
2. 当前结点的右儿子结点为空时出栈

抱歉又说了废话…其实也不是，每次 push 后第一个 pop 时肯定代表着当前结点的左儿子为空，其后的 pop 代表着当前右儿子为空（想一想为什么？）
大概轮廓有了，剩下的就是细节了，最麻烦的一个就是，如何找到"当前结点"
当 pop 时，出栈的栈顶结点肯定是"当前结点"了，
而当 push 时，入栈的结点肯定是"当前结点"啦

还原出了树用递归实现后序遍历，（毕竟非递归有点麻烦）

#include<iostream>
#include<stack>
#include<string>
#include<stdio.h>
#include<malloc.h>
using namespace std;
typedef struct TreeNode *Tree;
struct TreeNode{
	string data;
	Tree left;   // 左子树 
	Tree right;  // 右子树 
};
// 初始化一个树结点 
Tree create(){
	Tree T;
	T = (Tree)malloc(sizeof(struct TreeNode));
	T->left = NULL;
	T->right = NULL;
	return T; 
}

// 根据中序遍历整理出这棵树 
Tree restore(Tree T){
	int n;
	string str;
	stack<Tree> s;
	Tree node = T;
	bool flag = false;
	string value;
	scanf("%d\n",&n);
	// 根节点赋值 
	getline(cin,str);
	value = str.substr(5);  // 从第五个开始截取
	node->data = value;
	// 根结点入栈 
	s.push(node); 
	for(int i=1;i<2*n;i++){
		getline(cin,str);
		if(str=="Pop"){// 如果是 pop 操作
			node = s.top();
			s.pop(); 
		}else{   // push
			value = str.substr(5);  // 从第五个开始截取
			Tree tmp = create();
			tmp->data = value;
			if(!node->left){// 如果左儿子空，新结点就是左儿子 
				node->left = tmp;
				node = node->left; 
			}else if(!node->right){  // 如果右儿子空，新结点就是右儿子 
				node->right = tmp;
				node = node->right;
			}
			s.push(tmp);
		}
	}
	return T;
}

// 后序递归遍历
void bl(Tree T,bool &flag){
	if(T){
		bl(T->left,flag);
		bl(T->right,flag);
		if(!flag)
			flag = true;
		else
			cout<<" ";
		cout<<T->data;
	}
} 
int main(){
	Tree T;
	bool flag = false;
	string str;
	T = create();
	T = restore(T);
	bl(T,flag);
	return 0;
}