数据结构 03-树3 Tree Traversals Again (25 分)
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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.
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
给定先序遍历和中序遍历生成二叉树 输出后续遍历的结果
#include <iostream> #include <vector> #include <map> using namespace std; class tnode{ public: int data; tnode* left{nullptr}; tnode* right{nullptr}; tnode()=default; tnode(int d):data{d}{ } }; class tree{ public: tnode* root; int height; tree(){ tnode* node=new tnode; root=node; } }; void generateTree(tree &t){ //根据给定的先序和中序遍历 //建树 int n,temp; vector<tnode*> stack; string str; cin >>n; cin >> str >> temp; tnode* curNode=t.root; curNode->data=temp; curNode->left=nullptr; curNode->right=nullptr; stack.push_back(curNode); for(int i=0;i<2*n;i++){ cin >> str; if(str=="Pop"){// curNode=stack.back(); stack.pop_back(); }else{ cin >> temp; tnode* newNode=new tnode{temp}; if(!curNode->left){ curNode->left=newNode; curNode=curNode->left; }else if(!curNode->right){ curNode->right=newNode; curNode=curNode->right; } stack.push_back(newNode); } } } void postOrderTraversal(tnode* p,int rootData){ if(p){ if(p->left){ postOrderTraversal(p->left,rootData); } if(p->right){ postOrderTraversal(p->right,rootData); } cout << p->data; if(rootData!=p->data){ cout<<" "; } } } int main(){ tree t; generateTree(t); postOrderTraversal(t.root,t.root->data); return 0; }
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