树的遍历 (和) 玩转二叉树 的总结博客

树的遍历：

 

关键在与如何利用中序和后序建树。

　　中序的作用，提供了左右子树的所有节点的分布！

　　

　　前序，后序的作用：就是第一个元素（最后一个元素）提供了根节点，让中序找到子树的分布情况。

 

 

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
const int maxn = 100;
int af[maxn], in[maxn], n;
struct node{
    int l, r;
}tree[maxn];

int biuld(int al, int ar, int bl, int br){
    if (al > ar)return 0;
    int root, p1, len;
    root = af[br];
    for (p1 = al; p1 <= ar; ++p1)if (in[p1] == root)break;
    len = p1 - al;
    tree[root].l = biuld(al, p1 - 1, bl, bl + len-1);
    tree[root].r = biuld(p1 + 1, ar, bl + len, br - 1);
    return root;
}

void BFS(int x){
    queue<int>q;
    vector<int>v;
    q.push(x);
    while (!q.empty()){
        int root = q.front();    q.pop();
        if (root == 0)break;
        v.push_back(root);
        if (tree[root].l)q.push(tree[root].l);
        if (tree[root].r)q.push(tree[root].r);
    }
    int len = v.size();
    for (int i = 0; i < len; ++i)
        cout << v[i] << " \n"[i == len - 1];
}

int main(){
    cin >> n;
    for (int i = 0; i < n; ++i)cin >> af[i];
    for (int i = 0; i < n; ++i)cin >> in[i];
    int root = biuld(0, n - 1, 0, n - 1);
    BFS(root);
}

 

 

玩转二叉树：

这里的改变只是在建树的时候，把原本左子树放在实际的右子树上，原本的右子树放在左子树上。

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
const int maxn = 1e3;
struct node{
    int l, r;
}tree[maxn];
int pre[maxn], in[maxn], n;

int build(int al, int ar, int bl, int br){
    if (al > ar)return 0;
    int root, p1, len;
    root = pre[bl];
    for (p1 = al; p1 <= ar; ++p1)if (in[p1] == root)break;
    len = p1 - al;
    tree[root].r = build(al, p1 - 1, bl + 1, bl + len );
    tree[root].l = build(p1 + 1, ar, bl + len + 1, br);
    return root;
}

void BFS(int x){
    queue<int>q;        vector<int>v;
    q.push(x);
    while (!q.empty()){
        int root = q.front();    q.pop();
        if (root == 0)break;
        v.push_back(root);
        if (tree[root].l)q.push(tree[root].l);
        if (tree[root].r)q.push(tree[root].r);
    }

    int size = v.size();
    for (int i = 0; i < size; ++i)
        cout << v[i] << " \n"[i == size - 1];
}

int main(){
    cin >> n;
    for (int i = 0; i < n; ++i)cin >> in[i];
    for (int i = 0; i < n; ++i)cin >> pre[i];
    int root = build(0, n - 1, 0, n - 1);
    BFS(root);

}