C++實現利用（前序和中序生成二叉樹）以及（二叉樹的鏡像）
阿新 • • 發佈：2018-04-15
lse pub 非遞歸 ace 方法 [] reorder spa push
#include<iostream> #include<string.h> #include<stack> using namespace std; typedef struct BTree { int val; struct BTree *left,*right; }BTree; /*二叉樹的類，包含著操作二叉樹的各種方法*/ class Tree { public: BTree *create_node(int level,string pos); void PreOrder(BTree *t); //先序遍歷 voidInOrder(BTree *t); //中序遍歷 void PostOrder(BTree *t); //後序遍歷 void NonRecursivePreOrder(BTree*t); //非遞歸前序遍歷 void NonRecursiveInOrder(BTree*t); //非遞歸中序遍歷 void NonRecursivePostOrder(BTree*t);//非遞歸後序遍歷 BTree* PreOrder_and_INorder_RemakeTree(int *startPreorder,int *endPreder,int *startInorder,int*endInorder); BTree *root; }; /*用先序遍歷的方法遞歸構造一課二叉樹*/ BTree* Tree::create_node(int level,string pos) { int data; BTree *node = new BTree; int a[]={100,99,98,97,0,0,20,0,0,10,0,0,5,1,0,0,2,0,0}; static int t=0; cout<<"please enter data:level "<<level<<" "<<pos<<"--->值為:"<<a[t]<<endl; data=a[t++]; if(data == 0) { return NULL; } node->val= data; node->left = create_node(level+1,"left"); node->right= create_node(level+1,"right"); return node; } void Tree::PreOrder(BTree *t) { if(t) { cout<<t->val<<" ";; PreOrder(t->left); PreOrder(t->right); } } void Tree::InOrder(BTree *t) { if(t) { InOrder(t->left); cout<<t->val<<" ";; InOrder(t->right); } } void Tree::PostOrder(BTree *t) { if(t) { PostOrder(t->left); PostOrder(t->right); cout<<t->val<<" "; } } void Tree::NonRecursivePreOrder(BTree*t) { if(t==NULL) return; stack<BTree*>s; BTree *p; p=t; while(p||!s.empty()) { if(p) { cout<<p->val<<" "; s.push(p); p=p->left; } else{ p=s.top(); p=p->right; s.pop(); } } } void Tree::NonRecursiveInOrder(BTree*t) { if(t==NULL) return; stack<BTree*>s; BTree*p; p=t; while(p||!s.empty()) { if(p) { s.push(p); p=p->left; } else { p=s.top(); cout<<p->val<<" "; p=p->right; s.pop(); } } } void Tree::NonRecursivePostOrder(BTree*t) { if(t==NULL) return; stack<BTree*>s; BTree*p=t; BTree*r; while(p||!s.empty()) { if(p) { s.push(p); p=p->left; } else { p=s.top(); if(p->right&&p->right!=r) { p=p->right; s.push(p); p=p->left; } else { cout<<p->val<<" "; r=p; s.pop(); p=NULL; } } } } BTree* Tree::PreOrder_and_INorder_RemakeTree(int *startPreorder,int *endPreorder,int *startInorder,int*endInorder) { int rootValue=startPreorder[0]; BTree*root=new BTree; root->val=rootValue; root->left=NULL; root->right=NULL; // 在中序遍歷中找根節點的值 int*rootInorder=startInorder; while(rootInorder<=endInorder&&*rootInorder!=rootValue) rootInorder++; int leftLength=rootInorder-startInorder; int *leftPreorderEnd=startPreorder+leftLength; if(leftLength>0) { root->left=PreOrder_and_INorder_RemakeTree(startPreorder+1,leftPreorderEnd,startInorder,rootInorder-1); } if(leftLength<endPreorder-startPreorder) { root->right=PreOrder_and_INorder_RemakeTree(leftPreorderEnd+1,endPreorder,rootInorder+1,endInorder); } return root; } BTree* binary_tree_mirror(BTree*head) { BTree*newHead=head; if(head==NULL) return NULL; if(head->left!=NULL&&head->right!=NULL) { BTree *p; p=head->left; head->left=head->right; head->right=p; } binary_tree_mirror(head->left); binary_tree_mirror(head->right); return newHead; } int main() { Tree tree; tree.root = tree.create_node(1,"root"); cout<<"Pre"<<endl; tree.PreOrder(tree.root); cout<<endl; cout<<"非遞歸前序遍歷"<<endl; tree.NonRecursivePreOrder(tree.root); cout<<endl; cout<<"In"<<endl; tree.InOrder(tree.root); cout<<endl; cout<<"非遞歸中序遍歷"<<endl; tree.NonRecursiveInOrder(tree.root); cout<<endl; cout<<"Post"<<endl; tree.PostOrder(tree.root); cout<<endl; cout<<"非遞歸後序遍歷"<<endl; tree.NonRecursivePostOrder(tree.root); int preNum[]={100,99,98,97,20,10,5,1,2}; int InNum[]={97,98,20,99,10,100,1,5,2}; BTree*root2; int *endPreorder=&preNum[8]; int *endInorder=&InNum[8]; root2=tree.PreOrder_and_INorder_RemakeTree(preNum,endPreorder,InNum,endInorder); cout<<endl; cout<<"用後序遍歷測試用前序和中序生成的二叉樹："<<endl; tree.PostOrder(root2); cout<<"二叉樹的鏡像為:"<<endl; BTree *newTree; newTree=binary_tree_mirror(root2); cout<<"鏡像二叉樹的後序遍歷為:"<<endl; tree.PostOrder(newTree); return 0; } 二叉樹的圖：
(100)
(99) (5)
(98) (10) (1) (2)
(97) (20)
結果：
please enter data:level 1 root--->值為:100 please enter data:level 2 left--->值為:99 please enter data:level 3 left--->值為:98 please enter data:level 4 left--->值為:97 please enter data:level 5 left--->值為:0 please enter data:level 5 right--->值為:0 please enter data:level 4 right--->值為:20 please enter data:level 5 left--->值為:0 please enter data:level 5 right--->值為:0 please enter data:level 3 right--->值為:10 please enter data:level 4 left--->值為:0 please enter data:level 4 right--->值為:0 please enter data:level 2 right--->值為:5 please enter data:level 3 left--->值為:1 please enter data:level 4 left--->值為:0 please enter data:level 4 right--->值為:0 please enter data:level 3 right--->值為:2 please enter data:level 4 left--->值為:0 please enter data:level 4 right--->值為:0 Pre 100 99 98 97 20 10 5 1 2 非遞歸前序遍歷 100 99 98 97 20 10 5 1 2 In 97 98 20 99 10 100 1 5 2 非遞歸中序遍歷 97 98 20 99 10 100 1 5 2 Post 97 20 98 10 99 1 2 5 100 非遞歸後序遍歷 97 20 98 10 99 1 2 5 100 用後序遍歷測試用前序和中序生成的二叉樹： 97 20 98 10 99 1 2 5 100 鏡像二叉樹的後序遍歷為: 2 1 5 10 20 97 98 99 100
C++實現利用（前序和中序生成二叉樹）以及（二叉樹的鏡像）