1.给出中序遍历和后序遍历输出先序遍历
我们是怎么完成这个操作的呢？
主要的思想是：通过后序遍历找到根节点，通过中序遍历把左右子树分开，并记录左子树中节点的个数。通过记录的左子树节点的个数，准确找到左子树、右子树在中序、后序遍历中的位置。
in[]中序遍历的数组
post[]后序遍历的数组
在后序遍历中其最后一个节点为根节点，记为root
在中序遍历中找到一个节点值与root相同，即为下标i，这样就可以将树分为左右两部分了。
用numLeft标记左子树中节点的个数，左子树的节点个数为numLeft=i-inL;
建树的函数：
create(int postL,int postR,int inL,int inR);
postL是在后序遍历中左端点的位置，postR是后序遍历中右端点的位置；
inL是在中序遍历中左端点的位置，inR是中序遍历中右端点的位置。
对于左子树 传参 (postL,postL+numLeft-1,inL,i-1)
对于右子树 传参 (postL+numLeft,postR-1,i+1,inR)
1 #include <unordered_map> 2 #include <vector> 3 #include <iostream> 4 #include <algorithm> 5 using namespace std; 6 int in[100],post[100],n; 7 struct node 8 { 9 int data; 10 node* left; 11 node* right; 12 }; 13 node* create(int postL,int postR,int inL,int inR){ 14 if(postL>postR)return NULL; 15 node* root=new node; 16 root->data=post[postR]; 17 int i; 18 for(i=inL;i<=inR;i++){//在中序遍历的正确范围中找与根节点相同的值 19 if(in[i]==post[postR])break; 20 } 21 int numLeft; 22 numLeft=i-inL; 23 root->left=create(postL,postL+numLeft-1,inL,i-1); 24 root->right=create(postL+numLeft,postR-1,i+1,inR); 25 return root; 26 } 27 void pre(node* root) 28 { 29 cout<<root->data<<" "; 30 if(root->left!=NULL)pre(root->left); 31 if(root->right!=NULL)pre(root->right); 32 } 33 int main() 34 { 35 cin>>n; 36 int i; 37 for (int i = 0; i < n; ++i) { 38 cin>>in[i]; 39 } 40 for (int i = 0; i < n; ++i) { 41 cin>>post[i]; 42 } 43 node* root=create(0,n-1,0,n-1); 44 pre(root); 45 return 0; 46 }
层次遍历
void order(node* t) { if(!t)return; node* temp; queue<node*>q; q.push(t); while (!q.empty()) { temp=q.front(); a.push_back(temp->data); if(temp->left)q.push(temp->left); if(temp->right)q.push(temp->right); q.pop(); } }
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2
题意很简单，就是给出后序中序按层次遍历输出。
1 #include <unordered_map> 2 #include <vector> 3 #include <iostream> 4 #include <algorithm> 5 #include<queue> 6 using namespace std; 7 int in[100],post[100],n; 8 struct node 9 { 10 int data; 11 node* left; 12 node* right; 13 }; 14 node* create(int postL,int postR,int inL,int inR) 15 { 16 if(postL>postR)return NULL; 17 node* root=new node; 18 root->data=post[postR]; 19 int i; 20 for(i=inL;i<=inR;i++){ 21 //if(in[i]==root->data) 22 if(in[i]==post[postR]) 23 break; 24 } 25 int numLeft=i-inL; 26 root->left=create(postL,postL+numLeft-1,inL,i-1); 27 root->right=create(postL+numLeft,postR-1,i+1,inR); 28 return root; 29 } 30 vector<int>a; 31 void order(node* t) 32 { 33 if(!t)return; 34 node* temp; 35 queue<node*>q; 36 q.push(t); 37 while (!q.empty()) 38 { 39 temp=q.front(); 40 a.push_back(temp->data); 41 if(temp->left)q.push(temp->left); 42 if(temp->right)q.push(temp->right); 43 q.pop(); 44 } 45 } 46 int main() 47 { 48 cin>>n; 49 for(int i=0;i<n;i++) 50 cin>>post[i]; 51 for(int i=0;i<n;i++) 52 cin>>in[i]; 53 node* root=create(0,n-1,0,n-1); 54 order(root); 55 for(int i=0;i<a.size();i++) 56 { 57 if(i==0)cout<<a[i]; 58 else cout<<" "<<a[i]; 59 } 60 return 0; 61 }