二叉樹C++實現資料結構實驗__亂碼問題

#include <iostream>#include <string.h>#include <stack>#include <queue>using namespace std;template<class T>struct BiNode//二叉數節點{T data;BiNode<T>* lchild, *rchild;};template<class T>//模板類class BiTree{public:BiTree();//預設建構函式~BiTree();//解構函式BiNode<T>* GetRoot();//返回根節點void PreOrder(BiNode<T>* node);//先序遍曆void InOrder(BiNode<T>* node);//中序遍曆void PostOrder(BiNode<T>* node);//後序遍曆void LevelOrder(BiNode<T>* node);//層次遍曆//選做非遞迴實現void PreOrderNonRec(BiNode<T>* node);//先void InOrderNonRec(BiNode<T>* node);//中void PostOrderNonRec(BiNode<T>* node);//後//選做深度,節點數,葉子節點數,int NodesNum(BiNode<T>* node);//節點數int TreeDepth(BiNode<T>* node);//深度int LeafNum(BiNode<T>* node);//葉子節點數//選做交換子樹void SwapChild(BiNode<T>* node);//交換子樹private:BiNode<T>* m_root;//擷取根節點BiNode<T>* Create();//建立二叉樹};template<class T>BiTree<T>::BiTree(){m_root = new BiNode<T>;m_root = Create();}template<class T>BiTree<T>::~BiTree(){}template<class T>BiNode<T>* BiTree<T>::Create()//1. 按先序序列構造一棵二叉鏈表表示的二叉樹T；{char ch=getchar();BiNode<T>* pnode;if (ch == ' ')pnode = NULL;else{pnode = new BiNode<T>;pnode->data = ch;pnode->lchild = Create();pnode->rchild = Create();}return pnode;}template<class T>BiNode<T>* BiTree<T>::GetRoot(){return m_root;}template<class T>void BiTree<T>::PreOrder(BiNode<T>* node){if (!node)return;else{cout << node->data;PreOrder(node->lchild);PreOrder(node->rchild);}}template<class T>void BiTree<T>::InOrder(BiNode<T>* node){//前後定根，中序定左右if (!node)return;else{InOrder(node->lchild);cout << node->data;InOrder(node->rchild);}}template<class T>void BiTree<T>::PostOrder(BiNode<T>* node){if (!node)return;else{PostOrder(node->lchild);PostOrder(node->rchild);cout << node->data;}}template<class T>void BiTree<T>::LevelOrder(BiNode<T>* node){//層次遍曆需要queue來實現，思路：//@1初始化queue//if root為空白 返回//@2 push(root)//@3 while(queue不為空白)//s <-- queue.front()//queue.pop()//輸入s.data//if(s的左子樹不空)//s的左子樹入隊//if(s的右子樹不空)//s的右子樹入隊queue<BiNode<T>*> q;BiNode<T>* s = node;if (!s)return;q.push(s);while (!q.empty()){s = q.front();q.pop();cout << s->data;if (s->lchild)q.push(s->lchild);if (s->rchild)q.push(s->rchild);}}//先序遍曆非遞迴需要藉助stack s來實現，類比遞迴調用//總的迴圈邊界是當前節點不為空白或者stack不空，template<class T>void BiTree<T>::PreOrderNonRec(BiNode<T>* node){stack<BiNode<T>*> s;BiNode<T>* p = node;while (p|| !s.empty()){/*@1.每次找當前的節點的左子節點直到左為空白,經過的節點入棧,@2.然後彈出當前節點,搜尋一次右節點,如果p為空白並且s空則退出否則繼續@1*/while (p)//這裡執行VL{cout << p->data;//Vs.push(p);//訪問過的加入棧p = p->lchild;//L}if (!s.empty())//這裡執行R{p = s.top();s.pop();p = p->rchild;//R}}}template<class T>void BiTree<T>::InOrderNonRec(BiNode<T>* node){//@1 在當前節點p非空時候，將p入棧s,p的左子樹賦給p,保證左子樹都能入棧//p為空白時候，也就是左子樹最左邊訪問到了，這時候在棧非空的時候//@2 取棧頂給p，輸入p，出棧，這時候最底層的最左邊節點訪問了，將p的右子樹賦給p，重複@1stack<BiNode<T>*> s;BiNode<T>* p = node;while (p|| !s.empty()){while (p)//這裡執行L{s.push(p);p = p->lchild;}if (!s.empty())//這裡執行VR{p = s.top();cout << p->data;s.pop();p = p->rchild;}}}template<class T>void BiTree<T>::PostOrderNonRec(BiNode<T>* node){//訪問子節點的條件有兩種//1.當前節點的左右節點都為空白,可以直接存取//2.前一個被訪問的節點是當前節點的子節點//這樣就需要兩個指標,一個指向當前一個指向前一個被訪問的節點//然後保證入棧順序是先右再左,(這裡先壓右再壓左,這樣左在上面,就先訪問左)if (!node)return;stack<BiNode<T>*> s;s.push(node);BiNode<T>* pre = NULL;BiNode<T>* cur;while (!s.empty()){cur = s.top();if (!cur->lchild&& !cur->rchild ||(pre != NULL) && (pre == cur->lchild || pre == cur->rchild))//上一次訪問的是當前節點的左子樹{cout << cur->data;s.pop();pre = cur;//pre是前一個被訪問的節點}else{if (cur->rchild)s.push(cur->rchild);if (cur->lchild)s.push(cur->lchild);}}}template<class T>int BiTree<T>::LeafNum(BiNode<T>* node){//遞迴思路:找到葉子節點傳回值加一,傳回值計數//2種情況//1.節點為空白,返回 0,傳遞迴去//2.每當到達葉子節點,返回 1,傳遞給上一層函數if (!node)return 0;if (!node->lchild&&!node->rchild)return 1;return LeafNum(node->lchild) + LeafNum(node->rchild);}template<class T>int BiTree<T>::TreeDepth(BiNode<T>* node){/*遞迴思路:每個節點都有自己的左右子樹,每次返回當前節點左右子樹長度大的那個1.如果根節點為空白，則深度為0，返回0，遞迴的出口2.否則深度至少為1，然後累加他們左右子樹的深度，*/int LChildDep = 1, RChildDep = 1;if (!node)return 0;LChildDep += TreeDepth(node->lchild);//每次返回之前子樹的長度RChildDep += TreeDepth(node->rchild);return (LChildDep>RChildDep) ? (LChildDep) : (RChildDep);}template<class T>int BiTree<T>::NodesNum(BiNode<T>* node){//思路,遞迴遍曆所有節點,如果不是空節點的話,遞迴傳回值加1if (!node)return 0;return NodesNum(node->lchild) + NodesNum(node->rchild) + 1;}template<class T>void BiTree<T>::SwapChild(BiNode<T>* node){//思路,交換所有節點的節點,每個節點走一遍if (node){swap(node->lchild, node->rchild);SwapChild(node->lchild);SwapChild(node->rchild);}}/*範例輸入:ABC  DE G  F    範例輸出:先序建立一棵二叉樹,請輸入節點的值:ABC  DE G  F   建立完畢.先序:ABCDEGF中序:CBEGDFA後序:CGEFDBA層序:ABCDEFG選做題1:採用非遞迴演算法實現二叉樹遍曆先序:ABCDEGF中序:CBEGDFA後序:CGEFDBA選做題2:求二叉樹的深度/結點數目/葉結點數目TreeDepth is 5NodesNum is 7LeafNum is 3選做題3:將二叉樹每個結點的左右子樹交換位置。交換完畢.先序:ABDFEGC中序:AFDGEBC後序:FGEDCBA層序:ABDCFEG*/int main(){cout << "先序建立一棵二叉樹,請輸入節點的值:";BiTree<char> bitree;cout << "建立完畢." << endl;cout << "先序:";bitree.PreOrder(bitree.GetRoot());cout << endl;cout << "中序:";bitree.InOrder(bitree.GetRoot());cout << endl;cout << "後序:";bitree.PostOrder(bitree.GetRoot());cout << endl;cout << "層序:";bitree.LevelOrder(bitree.GetRoot());cout << endl;cout << "選做題1:採用非遞迴演算法實現二叉樹遍曆" << endl;cout << "先序:";bitree.PreOrderNonRec(bitree.GetRoot());cout << endl;cout << "中序:";bitree.InOrderNonRec(bitree.GetRoot());cout << endl;cout << "後序:";bitree.PostOrderNonRec(bitree.GetRoot());cout << endl;cout << "選做題2:求二叉樹的深度/結點數目/葉結點數目" << endl;cout << "深度為    : " << bitree.TreeDepth(bitree.GetRoot()) << endl;cout << "結點數目  : " << bitree.NodesNum(bitree.GetRoot()) << endl;cout << "葉結點數目: " << bitree.LeafNum(bitree.GetRoot()) << endl;cout << "選做題3:將二叉樹每個結點的左右子樹交換位置。" << endl;bitree.SwapChild(bitree.GetRoot());cout << "交換完畢." << endl;cout << "先序:";bitree.PreOrder(bitree.GetRoot());cout << endl;cout << "中序:";bitree.InOrder(bitree.GetRoot());cout << endl;cout << "後序:";bitree.PostOrder(bitree.GetRoot());cout << endl;cout << "層序:";bitree.LevelOrder(bitree.GetRoot());cout << endl;system("pause");}