1. 二叉樹的遍歷：先序（遞歸、非遞歸），中序（遞歸、非遞歸），后序（遞歸、非遞歸）。

#include <iostream> #include <string> #include <stack>using namespace std;struct BiTree {int NodeData = 0;struct BiTree *pLeft = nullptr;struct BiTree *pRight = nullptr; };//遍歷二叉樹： void show(struct BiTree *pRoot, int n) {if (pRoot == nullptr){return;}else{show(pRoot->pLeft, n + 1);for (int i = 0; i < n; i++)cout << " ";cout << pRoot->NodeData << endl;show(pRoot->pRight, n + 1);}} //-------------------------------------------------------------- //遞歸中序遍歷： void RecMidTravel(struct BiTree *pRoot) {if (pRoot == nullptr){return;}else {if (pRoot->pLeft != nullptr){RecMidTravel(pRoot->pLeft);}cout << pRoot->NodeData << endl;if (pRoot->pRight != nullptr){RecMidTravel(pRoot->pRight);}} } //中序非遞歸 void MidTravel(struct BiTree *pRoot) {if (pRoot == nullptr){return;}else{struct BiTree *pcur = pRoot;stack<BiTree *> mystack;while (!mystack.empty() || pcur != nullptr){while (pcur != nullptr){mystack.push(pcur);pcur = pcur->pLeft; //左節(jié)點(diǎn)全部進(jìn)棧 }if (!mystack.empty()){pcur = mystack.top();cout << pcur->NodeData << endl;mystack.pop(); //出棧pcur = pcur->pRight; //右節(jié)點(diǎn) }}} } //-------------------------------------------------------------- //遞歸先序遍歷： void RecPreTravel(struct BiTree *pRoot) {if (pRoot == nullptr){return;}else{cout << pRoot->NodeData << endl;if (pRoot->pLeft != nullptr){RecPreTravel(pRoot->pLeft);}if (pRoot->pRight != nullptr){RecPreTravel(pRoot->pRight);}} } //先序非遞歸 void PreTravel(struct BiTree *pRoot) {if (pRoot == nullptr){return;}else{struct BiTree *pcur = pRoot;stack<BiTree *> mystack;while (!mystack.empty() || pcur != nullptr){while (pcur != nullptr){cout << pcur->NodeData << endl;mystack.push(pcur);pcur = pcur->pLeft; //左節(jié)點(diǎn)全部進(jìn)棧 }if (!mystack.empty()){pcur = mystack.top();mystack.pop(); //出棧pcur = pcur->pRight; //右節(jié)點(diǎn) }}} }//-------------------------------------------------------------- //遞歸后序遍歷： void RecPostTravel(struct BiTree *pRoot) {if (pRoot == nullptr){return;}else{if (pRoot->pLeft != nullptr){RecPostTravel(pRoot->pLeft);}if (pRoot->pRight != nullptr){RecPostTravel(pRoot->pRight);}cout << pRoot->NodeData << endl;} } //后序非遞歸 struct nosame //標(biāo)識節(jié)點(diǎn)是否反復(fù)出現(xiàn) {struct BiTree *pnode;bool issame; };void PostTravel(struct BiTree *pRoot) {if (pRoot == nullptr){return;}else{struct BiTree *pcur = pRoot;stack<nosame *> mystack; //避免重復(fù)出現(xiàn)nosame *ptemp;while (!mystack.empty() || pcur != nullptr){while (pcur != nullptr){nosame *ptr = new nosame;ptr->issame = true;ptr->pnode = pcur;//節(jié)點(diǎn)//cout << pcur->NodeData << endl; mystack.push(ptr);pcur = pcur->pLeft; //左節(jié)點(diǎn)全部進(jìn)棧 }if (!mystack.empty()){ptemp = mystack.top();mystack.pop(); //出棧if (ptemp->issame == true) //第一次出現(xiàn) {ptemp->issame = false;mystack.push(ptemp);pcur = ptemp->pnode->pRight;//跳到右節(jié)點(diǎn) }else{cout << ptemp->pnode->NodeData << endl;//打印數(shù)據(jù)pcur = nullptr;}}}} }void main() {struct BiTree *pRoot;struct BiTree node1;struct BiTree node2;struct BiTree node3;struct BiTree node4;struct BiTree node5;struct BiTree node6;struct BiTree node7;struct BiTree node8;node1.NodeData = 1;node2.NodeData = 2;node3.NodeData = 3;node4.NodeData = 4;node5.NodeData = 5;node6.NodeData = 6;node7.NodeData = 7;node8.NodeData = 8;pRoot = &node1;node1.pLeft = &node2;node1.pRight = &node3;node2.pLeft = &node4;node2.pRight = &node5;node3.pLeft = &node6;node3.pRight = &node7;node4.pLeft = &node8;show(pRoot, 1);cout << "中序遞歸：" << endl;RecMidTravel(pRoot); //中序遞歸cout << "中序非遞歸：" << endl;MidTravel(pRoot); //中序非遞歸 cout << "先序遞歸：" << endl;RecPreTravel(pRoot);cout << "先序非遞歸：" << endl;PreTravel(pRoot); //先序非遞歸 cout << "后序遞歸：" << endl;RecPostTravel(pRoot);cout << "后序非遞歸：" << endl;PostTravel(pRoot); //后序非遞歸 cin.get(); }

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?2. 獲取二叉樹節(jié)點(diǎn)個(gè)數(shù)：

//遞歸獲取二叉樹節(jié)點(diǎn)個(gè)數(shù) int getNodeCount(BiTree *pRoot) {if (pRoot == nullptr){return 0;}else{return getNodeCount(pRoot->pLeft) + getNodeCount(pRoot->pRight) + 1;} }

　　　　

3. 判斷二叉樹是否為完全二叉樹：

//判斷二叉樹是否為完全二叉樹 bool isCompleteBiTree(BiTree *pRoot) {if (pRoot == nullptr){return false;}else{queue<BiTree *> myq;myq.push(pRoot);bool mustHaveChild = false; //必須有子節(jié)點(diǎn)bool result = true; //結(jié)果while (!myq.empty()){BiTree *node = myq.front();//頭結(jié)點(diǎn)myq.pop(); //出隊(duì)if (mustHaveChild) //必須有孩子 {if (node->pLeft != nullptr || node->pRight != nullptr){result = false;break;}} else{if (node->pLeft != nullptr && node->pRight != nullptr){myq.push(node->pLeft);myq.push(node->pRight);}else if (node->pLeft != nullptr && node->pRight == nullptr){mustHaveChild = true;myq.push(node->pLeft);}else if (node->pLeft == nullptr && node->pRight != nullptr){result = false;break;}else{mustHaveChild = true;}}}return result;} }

　　　　

4. 求二叉樹兩個(gè)節(jié)點(diǎn)的最小公共祖先：

//求二叉樹兩個(gè)節(jié)點(diǎn)的最小公共祖先 bool findnode(BiTree *pRoot, BiTree *node) //判斷節(jié)點(diǎn)是否在某個(gè)節(jié)點(diǎn)下 {if (pRoot == nullptr || node == nullptr){return false;}if (pRoot == node){return true;}bool isfind = findnode(pRoot->pLeft, node);if (!isfind){isfind = findnode(pRoot->pRight, node);}return isfind; }BiTree *getParent(BiTree *pRoot, BiTree *pChild1, BiTree *pChild2) {if (pRoot == pChild1 || pRoot == pChild2){return pRoot;}if (findnode(pRoot->pLeft, pChild1)){if (findnode(pRoot->pRight, pChild2)){return pRoot;} else{return getParent(pRoot->pLeft, pChild1, pChild2);}} else{if (findnode(pRoot->pLeft, pChild2)){return pRoot;}else{return getParent(pRoot->pRight, pChild1, pChild2);}} }

　　　　

5. 二叉樹的翻轉(zhuǎn)：

//二叉樹的翻轉(zhuǎn) BiTree *revBiTree(BiTree *pRoot) {if (pRoot==nullptr){return nullptr;}BiTree *leftp = revBiTree(pRoot->pLeft);BiTree *rightp = revBiTree(pRoot->pRight);pRoot->pLeft = rightp;pRoot->pRight = leftp; //交換return pRoot; }

　　　　

6. 求二叉樹第k層的節(jié)點(diǎn)個(gè)數(shù)：

//求二叉樹第K層的節(jié)點(diǎn)個(gè)數(shù) int getLevelConut(BiTree *pRoot, int k) {if (pRoot == nullptr || k < 1){return 0;}if (k == 1){return 1;}else{int left = getLevelConut(pRoot->pLeft, k - 1);int right = getLevelConut(pRoot->pRight, k - 1);return (left + right);} }

　　　　

7. 求二叉樹中節(jié)點(diǎn)的最大距離（相距最遠(yuǎn)的兩個(gè)節(jié)點(diǎn)之間的距離）：

//求二叉樹中節(jié)點(diǎn)的最大距離 struct res //用以遞歸間傳遞距離 {int maxDistance = 0;int maxDepth = 0; };res getMaxDistance(BiTree *pRoot) {if (pRoot == nullptr){res r1;return r1;}res leftr = getMaxDistance(pRoot->pLeft);res rightr = getMaxDistance(pRoot->pRight);res last; //最終結(jié)果last.maxDepth = max(leftr.maxDepth + 1, rightr.maxDepth + 1);//求最大深度last.maxDistance = max(max(leftr.maxDistance, rightr.maxDistance), leftr.maxDepth + rightr.maxDepth + 2);//求最大距離return last; }

　　　　

8. 判斷二叉樹是否為平衡二叉樹：

//判斷二叉樹是否為平衡二叉樹： bool isAVL(BiTree *pRoot, int & depth) //需要引用來傳遞數(shù)據(jù) {if (pRoot == nullptr){depth = 0;return true;}int leftdepth = 0;int rightdepth = 0;bool left = isAVL(pRoot->pLeft, leftdepth);bool right = isAVL(pRoot->pRight, rightdepth);if (left && right && abs(leftdepth - rightdepth) <= 1){depth = 1 + (leftdepth > rightdepth ? leftdepth : rightdepth);//深度return true;}else{return false;} }

　　　　　　　　

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轉(zhuǎn)載于:https://www.cnblogs.com/si-lei/p/9547016.html

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