《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.

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：

入栈当前结点的左儿子结点
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：

当前结点的左儿子结点为空时出栈
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;
}