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P272_P273.h
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P272_P273.h
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#pragma once
/**************************************************
@Author: Jie Feng
@Time: 2018/11/2
@File: P272_P273.h
@Description: The seventh week's assignment of data structure of Mr. Jianhua Lv
**************************************************/
#include <iostream>
#include <stack>
using namespace std;
template<class T> class Tree;
template<class T> class InorderIterator;
template<class T>
class TreeNode
{
friend class Tree<T>;
friend class InorderIterator<T>;
public:
TreeNode(T data)
{
this->data = data;
leftChild = 0;
rightChild = 0;
}
TreeNode<T> * GetLeftChild()
{
return leftChild;
}
TreeNode<T> * GetRightChild()
{
return rightChild;
}
private:
T data;
TreeNode<T> *leftChild;
TreeNode<T> *rightChild;
};
template<class T>
class Tree
{
public:
Tree(T node)
{
root = new TreeNode<T>(node);
}
~Tree();
TreeNode<T> * GetRoot()
{
return root;
}
void InsertToLeft(TreeNode<T> *parent, T leftChild);
void InsertToRight(TreeNode<T> *parent, T rightChild);
int CountLeafNode();
private:
TreeNode<T> *root;
};
template<class T>
void Tree<T>::InsertToLeft(TreeNode<T> *parent, T leftChild)
{
if (!parent->leftChild)
{
parent->leftChild = new TreeNode<T>(leftChild);
}
else
{
parent->leftChild->data = leftChild;
}
}
template<class T>
void Tree<T>::InsertToRight(TreeNode<T> *parent, T rightChild)
{
if (!parent->rightChild)
{
parent->rightChild = new TreeNode<T>(rightChild);
}
else
{
parent->rightChild->data = rightChild;
}
}
/*
以下对应于英文书P272 1
答:时间复杂度分析:因为采用的是二叉树的后序遍历思路来实现的叶子节点数目统计,
故每个节点都需要搜索一次,故时间复杂度为O(n),n为二叉树中节点个数
*/
template<class T>
int Tree<T>::CountLeafNode()
{
int numOfLeaf = 0;
stack<TreeNode<T> *> s;
TreeNode<T> *currNode = this->GetRoot();
TreeNode<T> *lastVisit = this->GetRoot();
while (currNode != 0 || !s.empty())
{
while (currNode)
{
s.push(currNode);
currNode = currNode->leftChild;
}
currNode = s.top();
if (currNode->rightChild == 0 || lastVisit == currNode->rightChild)
{
if (currNode->rightChild == 0 && currNode->leftChild == 0)
{
numOfLeaf++;
}
s.pop();
lastVisit = currNode;
currNode = 0;
}
else
{
currNode = currNode->rightChild;
}
}
return numOfLeaf;
}
/*
以下对应于英文书P273 4
答:时间复杂度分析:因为采用的是二叉树的中序遍历思路来实现从左至右依次删除节点,
故每个节点都需要入栈、出栈、然后被删除,故时间复杂度为O(n),n为二叉树中节点个数
*/
template<class T>
Tree<T>::~Tree()
{
stack<TreeNode<T> *> s;
TreeNode<T> *currNode = this->GetRoot();
TreeNode<T> *deleteNode = this->GetRoot();
//采用中序遍历的算法来依次从左至右删除节点
while (currNode != 0 || !s.empty())
{
while (currNode)
{
s.push(currNode);
currNode = currNode->leftChild;
}
if (!s.empty())
{
currNode = s.top();
deleteNode = currNode;
s.pop();
currNode = currNode->rightChild;
delete deleteNode;
deleteNode = 0;
}
}
}
void run_P272_P273()
{
Tree<char> binaryTree('A');
binaryTree.InsertToLeft(binaryTree.GetRoot(), 'B');
binaryTree.InsertToRight(binaryTree.GetRoot(), 'C');
binaryTree.InsertToLeft(binaryTree.GetRoot()->GetLeftChild(), 'D');
binaryTree.InsertToRight(binaryTree.GetRoot()->GetLeftChild(), 'E');
binaryTree.InsertToLeft(binaryTree.GetRoot()->GetRightChild(), 'F');
binaryTree.InsertToRight(binaryTree.GetRoot()->GetRightChild(), 'G');
binaryTree.InsertToLeft(binaryTree.GetRoot()->GetLeftChild()->GetLeftChild(), 'H');
//P272 1
cout << "二叉树中叶子节点数量为:" << binaryTree.CountLeafNode() << endl;
}