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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.
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
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <iostream> 4 #include <string.h> 5 #include <math.h> 6 #include <algorithm> 7 #include <string> 8 #include <stack> 9 #include <queue> 10 using namespace std; 11 const int maxn=33; 12 int n; 13 int pre[maxn],in[maxn]; 14 struct node{ 15 int data; 16 node *left; 17 node *right; 18 }; 19 //给一个先序和中序，求层次遍历 20 node *create(int a,int b,int c,int d) 21 { 22 if(a>b)return NULL; 23 node *temp=new node; 24 temp->data=pre[a]; 25 int k=0; 26 for(;k+c<=d;k++) 27 { 28 if(in[c+k]==pre[a])break; 29 } 30 temp->left=create(a+1,a+k,c,c+k-1); 31 temp->right=create(a+k+1,b,c+k+1,d); 32 return temp; 33 } 34 //后序 35 int printcount=0; 36 void postorder(node * root) 37 { 38 if(root==NULL)return; 39 postorder(root->left); 40 postorder(root->right); 41 printf("%d",root->data); 42 printcount++; 43 if(printcount<n)printf(" "); 44 } 45 void BFS(node * root) 46 { 47 int count=0; 48 queue<node*> q; 49 q.push(root); 50 while(!q.empty()) 51 { 52 node* now=q.front(); 53 q.pop(); 54 printf("%d",now->data); 55 count++; 56 if(count<n)printf(" "); 57 if(now->left!=NULL)q.push(now->left); 58 if(now->right!=NULL)q.push(now->right); 59 } 60 } 61 int main(){ 62 int temp,num=0,innum=0; 63 stack<int> st; 64 scanf("%d",&n); 65 for(int i=0;i<n*2;i++) 66 { 67 char a[5]; 68 scanf("%s",a); 69 if(strcmp(a,"Push")==0) 70 { 71 scanf("%d",&temp); 72 pre[num++]=temp; 73 st.push(temp); 74 }//pop 75 else 76 { 77 in[innum++]=st.top(); 78 st.pop(); 79 } 80 } 81 //根据前序中序 建树 82 83 node* root=create(0,n-1,0,n-1); 84 postorder(root); 85 return 0; 86 }
A1086. Tree Traversals Again (25)
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原文地址：http://www.cnblogs.com/ligen/p/4320746.html
