L2-006 树的遍历
给定一棵二叉树的后序遍历和中序遍历，请你输出其层序遍历的序列。这里假设键值都是互不相等的正整数。
输入格式：
输入第一行给出一个正整数N（<=30），是二叉树中结点的个数。第二行给出其后序遍历序列。第三行给出其中序遍历序列。数字间以空格分隔。
输出格式：
在一行中输出该树的层序遍历的序列。数字间以1个空格分隔，行首尾不得有多余空格。
输入样例：
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
输出样例：
4 1 6 3 5 7 2方法一：数组
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 35;
int n, in_order[MAXN], post_order[MAXN];
vector<int> ans;
struct Node{
int v;
int left, right;
}tree[MAXN];
int Root;
int build(int l1, int r1, int l2, int r2) {
if(l1 > r1) return -1;
int root = post_order[r2];
int p = l1;
while(in_order[p] != root) {
p++;
}
int cnt = p - l1;
tree[root].left = build(l1, p-1, l2, l2+cnt-1);
tree[root].right = build(p+1, r1, l2+cnt, r2-1);
return root;
}
void bfs() {
queue<int> q;
q.push(Root);
while(!q.empty()) {
int u = q.front();
q.pop();
ans.push_back(u);
if(tree[u].left != -1) {
q.push(tree[u].left);
}
if(tree[u].right != -1) {
q.push(tree[u].right);
}
}
}
int main() {
cin>>n;
for(int i = 0; i < n; i++) {
cin>>post_order[i];
}
for(int i = 0; i < n; i++) {
cin>>in_order[i];
}
Root = build(0, n-1, 0, n-1);
bfs();
for(int i = 0; i < ans.size(); i++) {
if(i != 0) {
cout<<" ";
}
cout<<ans[i];
}
cout<<endl;
return 0;
}
方法二指针：
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
const int MAXN = 35;
int n, in_order[MAXN], post_order[MAXN];
vector <int> ans;
struct Node{
int v;
Node *left, *right;
};
Node *Root;
Node * build(int l1, int r1, int l2, int r2) {
if(l1 > r1) return NULL;
Node *root = new Node;
root->v = post_order[r2];
int p = l1;
while(in_order[p] != root->v) {
p++;
}
int cnt = p - l1;
root->left = build(l1, p-1, l2, l2+cnt-1);
root->right = build(p+1, r1, l2+cnt, r2-1);
return root;
}
void bfs() {
queue<Node*> q;
q.push(Root);
while(!q.empty()) {
Node *u = q.front();
q.pop();
ans.push_back(u->v);
if(u->left != NULL) {
q.push(u->left);
}
if(u->right != NULL) {
q.push(u->right);
}
}
}
int main() {
cin>>n;
for(int i = 0; i < n; i++) {
cin>>post_order[i];
}
for(int i = 0; i < n; i++){
cin>>in_order[i];
}
Root = build(0, n-1, 0, n-1);
bfs();
for(int i = 0; i < ans.size(); i++) {
if(i != 0) {
cout<<" ";
}
cout<<ans[i];
}
cout<<endl;
return 0;
}