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.
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
核心思路
这部分的核心思路是push是先序遍历，pop是中序遍历，已知先序中序确定一棵树，然后后序在遍历树，如果不这样考虑，那就要从题目中本身出发，比如push代表它是根节点。
完整源码
#include<cstdio>
#include<cstring>
#include<stack>
#include<algorithm>
using namespace std;
const int maxn = 50;
struct node {
int data;
node *lchild;
node *rchild;
};
int pre[maxn],in[maxn],post[maxn]; //先序，中序，后序
int n ;//结点个数
//当前二叉树的先序序列为[preL,preR],中序序列为[inL,inR]
//create函数返回构建出的二叉树的根节点地址
node *create(int preL, int preR, int inL,int inR){
if(preL > preR){
return NULL;// 若先序序列长度小于等于0，则直接返回
}
node * root = new node;//新建一个新的结点，用来存放当前二叉树的根节点
root->data = pre[preL];//新结点的数据域为根节点的值
int k;
for(k = inL;k<=inR;k++){
if(in[k] == pre[preL]) break;
}
int numLeft = k - inL;//左子树的结点个数
//返回左子树的根节点地址，赋值给root的左指针
root->lchild = create(preL+1,preL+numLeft,inL,k-1);
root->rchild = create(preL+numLeft+1,preR,k+1,inR);
return root;//返回根节点的地址
}
int num = 0;//已经输出的结点个数
void postorder(node *root){
//后续遍历
if(root == NULL){
return ;
}
postorder(root->lchild);
postorder(root->rchild);
printf("%d",root->data);
num++;
if(num<n) printf(" ");
}
int main()
{
scanf("%d",&n);
char str[5];
stack<int>st;
int x,preIndex = 0,inIndex = 0;//入栈元素，先序序列位置及中序序列位置
for(int i =0;i<2*n;i++){//出栈入栈共2n次
scanf("%s",str);
if(strcmp(str,"Push") == 0){//入栈
scanf("%d",&x);
pre[preIndex++] = x; //令pre[preIndex] = x
st.push(x);
}else{
in[inIndex++] = st.top();// 令in[inIndex] = st.top
st.pop();
}
}
node *root = create(0,n-1,0,n-1);//建树
postorder(root);
return 0;
}
方法2，非栈模拟
//每到一棵树，第一个操作为push
//每当push ，必考察有没有左子树,看是不是push
//每当pop，考察有没有右子树,看是不是push
#include<iostream>
#include<vector>
using namespace std;
struct node{
int key;
node *left,*right;
};
vector<int>v;
void makenode(node *&root){
string s;
int i;
cin >> i;
root = new node;
root->key = i;
cin >> s;
if(s=="Push"){
makenode(root->left);
}else root->left = NULL;
s ="";
cin >> s;
if(s=="Push"){
makenode(root->right);
}else root->right = NULL;
}
void postordertraverse(node*root){
if(root==NULL) return;
postordertraverse(root->left);
postordertraverse(root->right);
v.emplace_back(root->key);
}
int main()
{
int i,j,k;
cin >> i;
node* tree = NULL;
string s;
cin>>s;
makenode(tree);
postordertraverse(tree);
for(i=0;i<v.size();i++){
if(i) cout << " ";
cout << v[i];
}
return 0;
}