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7-10 Tree Traversals Again(25 分)
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 (≤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
//出栈的顺序其实是中序遍历 输出后序遍历 #include<stdio.h> #include<string.h> #include<stack> using namespace std; int n; int pre[35],ld[35],rd[35]; int ans=0; void dfs(int u)//输出后序遍历 { if(ld[u]!=-1) dfs(ld[u]); if(rd[u]!=-1) dfs(rd[u]); if(ans==0) { printf("%d",u); ans=1; } else printf(" %d",u); } int main() { int i,u,v; char st[20]; stack<int>s; while(!s.empty()) { s.pop(); } scanf("%d",&n); scanf("%s",st); for(i=1;i<=n;i++) { pre[i]=i; ld[i]=rd[i]=-1; } if(strcmp(st,"Push")==0) { scanf("%d",&u); pre[u]=0;//根节点 s.push(u); } v=u; int nn=n*2-1; //printf("%d\n",nn); while(nn--) { scanf("%s",st); if(strcmp(st,"Push")==0) { scanf("%d",&u); //上一步操作的节点是栈顶元素则放入的是左孩子,否则是右孩子 if(!s.empty()&&v==s.top()) { pre[u]=v; ld[v]=u;//左孩子 } else { pre[u]=v; rd[v]=u; } s.push(u); v=s.top(); } else { v=s.top(); s.pop(); } } for(i=1;i<=n;i++) { if(pre[i]==0) { //dfs(i); break; } } //printf("%d\n",i); dfs(i); printf("\n"); return 0; }