二叉樹系列問題1

1、非遞迴實現二叉樹的三種遍曆

2、分層遍曆二叉樹

3、找出二叉樹上任意兩個節點的共同父節點

4、找出二叉樹中和為某值的所有路徑

5、二叉樹的鏡像

首先是樹節點和樹的類資料結構

#define ElementType intclass BinTree;class BinTreeNode{public:friend class BinTree;BinTreeNode():Left(NULL),Right(NULL){}BinTreeNode(ElementType temp, BinTreeNode * pLeft = NULL, BinTreeNode * pRight = NULL):data(temp), Left(pLeft), Right(pRight){}ElementType GetData() const;BinTreeNode * GetLeft() const;BinTreeNode *GetRight()const;void SetData(const ElementType temp);void SetLeft(BinTreeNode *pLeft);void SetRight(BinTreeNode *pRight);void InOrder();void PreOrder();void PostOrder();int Size();int Height();bool isAvlTree();//該節點是不是平衡的BinTreeNode *Copy(const BinTreeNode *copy);void Destroy();friend bool Equal(const BinTreeNode *s, const BinTreeNode *t);private:BinTreeNode *Left, *Right;ElementType data;};

class BinTree{public:BinTree():Root(NULL){}BinTree(const BinTree & temp);virtual ~BinTree(){Root->Destroy();}virtual bool IsEmpty();virtual BinTreeNode* GetLeft(BinTreeNode *current);virtual BinTreeNode* GetRight(BinTreeNode *current);virtual BinTreeNode* GetParent(BinTreeNode *current);const BinTreeNode* GetRoot()const;BinTreeNode* GetRoot();virtual bool Insert(const ElementType temp);virtual BinTreeNode *Find(const ElementType temp)const;void InOrder();void PreOrder();void PostOrder();void InOrderWithCycle();//迴圈實現樹的三種遍曆void PreOrderWithCycle();void PostOrderWithCycle();int Size();int Height();bool isAvlTree(); //判斷一課數是不是平衡二叉樹BinTree& operator=(const BinTree copy);friend bool operator == (const BinTree s, const BinTree t);friend ostream& operator<<(ostream& os, BinTree& temp);friend istream& operator>>(istream& is, BinTree& temp);private:BinTreeNode *Root;BinTreeNode * GetParent(BinTreeNode *start, BinTreeNode *current);BinTreeNode *Insert(BinTreeNode *, ElementType temp);void Print(BinTreeNode *start, int n = 0);};

1、非遞迴實現二叉樹的三種遍曆

void BinTree::InOrderWithCycle()//中{stack<BinTreeNode *> s;BinTreeNode *pMove;pMove = this->Root;while(pMove != NULL || !s.empty()){while(pMove != NULL){s.push(pMove);pMove = pMove->Left;}if(!s.empty()){pMove = s.top();s.pop();cout<<pMove->data<<" "<<endl;pMove = pMove->Right;}}}void BinTree::PreOrderWithCycle()//前{stack<BinTreeNode *> s;BinTreeNode *pMove;pMove = this->Root;while(pMove != NULL || !s.empty()){while(pMove != NULL){cout<<pMove->data <<" "<<endl;s.push(pMove);pMove = pMove->Left;}if(!s.empty()){pMove = s.top();s.pop();pMove = pMove->Right;}}}void BinTree::PostOrderWithCycle()//後{stack<BinTreeNode *> s;BinTreeNode *pMove;BinTreeNode *pre = NULL;pMove = this->Root;s.push(pMove);while(!s.empty()){pMove = s.top();if((pMove->Left == NULL && pMove->Right == NULL) || (pre != NULL && (pMove->Left == pre || pMove->Right == pre)))//判斷節點是否是葉子節點，或者該節點的左右子樹是否已經遍曆過{//pMove = s.top();cout<<pMove->data<<" "<<endl;pre = pMove;s.pop();}else{if(pMove->Right)s.push(pMove->Right);if(pMove->Left)s.push(pMove->Left);}}}

2、分層遍曆二叉樹

把節點入棧，然後左右節點入棧

void PrintTreeInlayer(BinTree &Tree)//分層遍曆二叉樹{queue<const BinTreeNode *> q;if(Tree.IsEmpty())return;const BinTreeNode *temp;q.push(Tree.GetRoot());while(!q.empty()){temp = q.front();cout<<temp->GetData()<<" "<<endl;q.pop();if(temp->GetLeft())q.push(temp->GetLeft());if(temp->GetRight())q.push(temp->GetRight());}}

3、找出二叉樹上任意兩個節點的最近共同父節點

分別遍曆兩個節點，並把經過的節點方入容器，然後從頭遍曆兩個容器，最後一個相同的節點就是最近的共同父節點

const BinTreeNode *FindPublicNode(ElementType first, ElementType second,const BinTreeNode *Root)//二叉樹的最近公用父節點{void FindPath(vector<const BinTreeNode *>&, ElementType ,const BinTreeNode *);vector<const BinTreeNode *>FirstVec, SecondVec;FindPath(FirstVec, first, Root);FindPath(SecondVec, second, Root);vector<const BinTreeNode *>::const_iterator Firstiter = FirstVec.begin();vector<const BinTreeNode *>::const_iterator Seconditer = SecondVec.begin();while(Firstiter != FirstVec.end() && Seconditer != SecondVec.end()){if(*Firstiter != *Seconditer)break;Firstiter++;Seconditer++;}if(Firstiter == FirstVec.end()){cout<<"First is the parent of second"<<endl;return *(--Firstiter);}if(Seconditer == SecondVec.end()){cout<<"Second is the parent of first"<<endl;return *(--Seconditer);}return *(--Firstiter);}void FindPath(vector<const BinTreeNode *>& s, ElementType data, const BinTreeNode *Root)//尋找節點，把路徑放入容器{const BinTreeNode *pMove(Root);while(pMove != NULL){if(data < pMove->GetData()){s.push_back(pMove);pMove = pMove->GetLeft();}else if(data > pMove->GetData()){s.push_back(pMove);pMove = pMove->GetRight();}else{s.push_back(pMove);break;}}}

4、找出二叉樹中和為某值的所有路徑

void FindSumPath(const BinTreeNode *Root, int sum, vector<ElementType>& path){bool isleaf(const BinTreeNode *);if(!Root)return;sum -= Root->GetData();path.push_back(Root->GetData());if(isleaf(Root) && sum == 0){for(vector<ElementType>::const_iterator iter = path.begin(); iter != path.end(); ++iter)cout<<*iter<<" ";cout<<endl;}if(Root->GetLeft())FindSumPath(Root->GetLeft(), sum, path);if(Root->GetRight())FindSumPath(Root->GetRight(), sum, path);sum += Root->GetData();path.pop_back();}

bool isleaf(const BinTreeNode *Root){return (Root->GetLeft()== NULL && Root->GetRight()== NULL) ? true : false;}

5、二叉樹的鏡像

BinTreeNode * TreeMirror(BinTreeNode *Root){if(Root == NULL)return NULL;BinTreeNode *temp = Root->GetRight();Root->SetRight(Root->GetLeft());Root->SetLeft(temp);BinTreeNode *Left = TreeMirror(Root->GetLeft());Root->SetLeft(Left);BinTreeNode *Right = TreeMirror(Root->GetRight());Root->SetRight(Right);return Root;}BinTreeNode *TreeMirror2(BinTreeNode *Root){if(Root == NULL)return NULL;else{TreeMirror2(Root->GetLeft());TreeMirror2(Root->GetRight());BinTreeNode *temp = Root->GetLeft();Root->SetLeft(Root->GetRight());Root->SetRight(temp);return Root;}}