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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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
#include<bits/stdc++.h> using namespace std; const int maxn=1010; #define inf 0x3fffffff stack<int> s; vector<int> pre; vector<int> mid; vector<int> post; void toPost(int root,int left,int right){//root为先序中根节点的下标,left和right分别为中序中子树的左右边界 if(left>right){ return ; } int i=left; while(i<right&&mid[i]!=pre[root]){ i++; } toPost(root+1,left,i-1); toPost(root+1+i-left,i+1,right); post.push_back(pre[root]); } int main(){ int n; scanf("%d",&n); int cnt=0; do{ string ss; getline(cin,ss); if(ss[1]=='u'){ ss=ss.substr(5); int ds=ss[0]-'0'; s.push(ds); pre.push_back(ds); } else if(ss[1]=='o'){ int k=s.top(); mid.push_back(k); s.pop(); cnt++; } }while(cnt<n); toPost(0,0,n-1); for(int i=0;i<n;i++){ if(i==n-1){ printf("%d ",post[i]); } else{ printf("%d ",post[i]); } } return 0; }
第五个测试点过不去,应该是输入的问题,另外,不太理解由先序和中序确定后序的代码逻辑。。。明天好好研究
修改输入方式后,成功AC:
#include<bits/stdc++.h> using namespace std; const int maxn=1010; #define inf 0x3fffffff stack<int> s; vector<int> pre; vector<int> mid; vector<int> post; //post(0,0,n-1); void toPost(int root,int left,int right){//root为先序中根节点的下标,left和right分别为中序中子树的左右边界 if(left>right){ return ; } int i=left; while(i<right&&mid[i]!=pre[root]){ i++; } toPost(root+1,left,i-1);//递归结束后,root到达左子树的左边叶子结点 toPost(root+1+i-left,i+1,right);//递归结束后,root到达右子树的右边叶子结点 post.push_back(pre[root]); } //pre(n-1,0,n-1); //void toPre(int root,int left,int right){ // if(left>right){ // return ; // } // int i=left; // while(i<right&&mid[i]==post[root]){ // i++; // } // toPre(root+1-right+i,left,i-1); // toPre(root-1,i+1,right); // pre.push_back(post[root]); //} int main(){ int n; scanf("%d",&n); int cnt=0; for(int i=0;i<2*n;i++){ string ss; cin>>ss; if(ss=="Push"){ int ds; cin>>ds; s.push(ds); pre.push_back(ds); } else { int k=s.top(); mid.push_back(k); s.pop(); } } toPost(0,0,n-1); for(int i=0;i<n;i++){ if(i==n-1){ printf("%d ",post[i]); } else{ printf("%d ",post[i]); } } return 0; }