《数据结构》03-树3 Tree Traversals Again

ゞ浴缸里的玫瑰 2022-05-15 15:58 184阅读 0赞

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.  
![20181030171007780.jpg][]  
 F i g u r e 1 Figure 1 Figure1  
**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;
    }

[20181030171007780.jpg]: /images/20220503/56eb31ff26f14fb2a9791aa495659dc1.png