1086 Tree Traversals Again (25 分)（树的遍历）

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

题目大意：

用栈的形式给出一棵二叉树的建立的顺序，求这棵二叉树的后序遍历

分析：

栈实现的是二叉树的中序遍历（左根右），而每次push入值的顺序是二叉树的前序遍历（根左右），所以该题可以用二叉树前序和中序转后序的方法做～

root为当前子树的根结点在前序pre中的下标，start和end为当前子树的最左边和最右边的结点在中序in中的下标。用i找到当前子树的根结点root在中序中的下标，然后左边和右边就分别为当前根结点root的左子树和右子树。递归实现～

Update：Github用户littlesevenmo给我发issue提出题目并没有说所有节点的值互不相同。因此，在有多个节点的值相同的情况下，之前的代码会输出错误的结果，所以修改后的代码中添加了key作为索引，前中后序中均保存索引值，然后用value存储具体的值，修改后的代码如下：

原文链接：https://blog.csdn.net/liuchuo/article/details/52181237

柳神题解

#include <cstdio>
#include <vector>
#include <stack>
#include <cstring>
using namespace std;
vector<int> pre, in, post,value;
void postorder(int root, int start, int end) {
    if (start > end) return;
    int i = start;
    while (i < end && in[i] != pre[root]) i++;
    postorder(root + 1, start, i - 1);
    postorder(root + 1 + i - start, i + 1, end);
    post.push_back(pre[root]);
}
int main() {
    int n;
    scanf("%d", &n);
    char str[5];
    stack<int> s;
    int key=0;
    while (~scanf("%s", str)) {
        if (strlen(str) == 4) {
            int num;
            scanf("%d", &num);
            value.push_back(num);
            pre.push_back(key);
            s.push(key++);
        } else {
            in.push_back(s.top());
            s.pop();
        }
    }
    postorder(0, 0, n - 1);
    printf("%d", value[post[0]]);
    for (int i = 1; i < n; i++)
        printf(" %d",value[post[i]]);
    return 0;
}

给出前序中序，重建二叉树

算法笔记题解

#include <bits/stdc++.h>

using namespace std;
const int maxn=50;
struct node
{
    int data;
    node* lchild;
    node* rchild;
};
int pre[maxn],in[maxn],post[maxn];
int n;

node* create(int preL,int preR,int inL,int inR)
{
    if(preL>preR) return nullptr;
    node* root=new node;
    root->data=pre[preL];
    int k;
    for(k=inL;k<=inR;k++){
        if(in[k]==pre[preL]) break;
    }
    int numLeft=k-inL;
    root->lchild=create(preL+1,preL+numLeft,inL,k-1);
    root->rchild=create(preL+numLeft+1,preR,k+1,inR);
    return root;
}
int num=0;
void postorder(node* root)
{
    if(root==nullptr) return;
    postorder(root->lchild);
    postorder(root->rchild);
    cout<<root->data;
    num++;
    if(num<n) cout<<" ";
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    cin>>n;
    char str[5];
    stack<int> st;
    int x,preInedx=0,inIndex=0;
    for(int i=0;i<2*n;i++){
        scanf("%s",str);
        if(strcmp(str,"Push")==0){
            cin>>x;
            pre[preInedx++]=x;
            st.push(x);
        }else{
            in[inIndex++]=st.top();
            st.pop();
        }
    }
    node* root=create(0,n-1,0,n-1);
    postorder(root);
    return 0;
}