剑指offer----面试题8 二叉树的下一个节点

具体题目看剑指OFFER书面试题8，本博文仅供博主代码笔记只用，如需转载请私信征得博主同意

本例主要需要学习：

1.输出层次分明的二叉树（先序遍历）

2.构建有父指针的二叉树的算法

3.销毁二叉树的算法

4.查找二叉树找那个任意指定的节点，按中序遍历的下一个节点

//#include <stdio.h>
//#include <stdlib.h>
#include <iostream>
using namespace std;

struct BinaryTreeNode
{
   int value;
   BinaryTreeNode* leftchild;
   BinaryTreeNode* rightchild;
   BinaryTreeNode* parent;
};

BinaryTreeNode* CreateTreeNode(int m_value)
{
   BinaryTreeNode* pNode = new BinaryTreeNode();
   pNode->value = m_value;
   pNode->leftchild = nullptr;
   pNode->rightchild = nullptr;
   pNode->parent = nullptr;

   return pNode;
}

void ConnectNode(BinaryTreeNode* Con_parent, BinaryTreeNode* Con_lchild,
   BinaryTreeNode* Con_rchild)
{
   Con_parent->leftchild = Con_lchild;
   Con_parent->rightchild = Con_rchild;
   Con_lchild->parent = Con_parent;
   Con_rchild->parent = Con_parent;
}

void DispBTNode(BinaryTreeNode* b)
{
   //若要层次更分明可考虑层次遍历输出，层次遍历代码参考我的根据中序先序重建二叉树的博文

//https://blog.csdn.net/qq_34793133/article/details/80582358

   cout << "二叉树输出如下，以括号标明层次：" << endl;
   if (b != nullptr)
   {
       cout << b->value;
   }

   if (b->leftchild != nullptr || b->rightchild != nullptr)
   {
       cout << '(';

       DispBTNode(b->leftchild);
       if (b->rightchild != nullptr)
           cout << ',' ;
       DispBTNode(b->rightchild);

       cout << ")";

   }

   cout << endl;
}

void DestroyBinaryTree(BinaryTreeNode* Root)
{
   if (Root != nullptr)
   {
       BinaryTreeNode* tmp_leftchild = Root->leftchild;
       BinaryTreeNode* tmp_rightchild = Root->rightchild;

       delete Root; Root = nullptr;

       DestroyBinaryTree(Root->leftchild);
       DestroyBinaryTree(Root->rightchild);
   }

}

BinaryTreeNode* GetNextNode(BinaryTreeNode* pNode)
{
   if (pNode == nullptr)
       return nullptr;

   if (pNode->rightchild != nullptr)
   {
       BinaryTreeNode* temp1 = pNode->rightchild;
       while (temp1->leftchild != nullptr)
       {
           temp1 = temp1->leftchild;
       }

       return temp1;
   }

   else if (pNode->parent != nullptr)
   {
       BinaryTreeNode* ParentNode = pNode->parent;
       BinaryTreeNode* CurrentNode = pNode;

       while (ParentNode != nullptr&&ParentNode->rightchild != CurrentNode)
       {
           CurrentNode = ParentNode; ParentNode = CurrentNode->parent;
       }

       return ParentNode;
   }
}

// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pNode, BinaryTreeNode* expected)
{
   if (testName != nullptr)
       printf("%s begins: ", testName);

   BinaryTreeNode* pNext = GetNextNode(pNode);
   if (pNext == expected)
       printf("Passed.\n");
   else
       printf("FAILED.\n");
}

//            8
//        6      10
//       5 7    9 11

int main(void)
{
   BinaryTreeNode* pNode8 = CreateTreeNode(8);
   BinaryTreeNode* pNode6 = CreateTreeNode(6);
   BinaryTreeNode* pNode10 = CreateTreeNode(10);
   BinaryTreeNode* pNode5 = CreateTreeNode(5);
   BinaryTreeNode* pNode7 = CreateTreeNode(7);
   BinaryTreeNode* pNode9 = CreateTreeNode(9);
   BinaryTreeNode* pNode11 = CreateTreeNode(11);

   ConnectNode(pNode8, pNode6, pNode10);
   ConnectNode(pNode6, pNode5, pNode7);
   ConnectNode(pNode10, pNode9, pNode11);

   DispBTNode(pNode8);

   Test("Test1", pNode8, pNode9);
   return 0;
}

剑指offer----面试题8 二叉树的下一个节点

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