每日一练(40) - 判断二叉树是否是一颗平衡二叉树

题目来自剑指Offer

题目：

思路：

类似后序遍历，同时要判断左右子树是否为平衡二叉树。

代码：

#include <iostream>
#include <assert.h>
using namespace std;

const int SIZE = 100;

struct BinaryTreeNode 
{
	char m_nValue;
	BinaryTreeNode* m_pLeft;
	BinaryTreeNode* m_pRight;
};

bool IsBalance(BinaryTreeNode* pRoot,int& nDepth)
{
	if (!pRoot)
	{
		nDepth = 0;
		return true;
	}
	int nLeftDepth = -1;
	int nRightDepth = -1;
	int nDiff = -1;
	if (IsBalance(pRoot->m_pLeft,nLeftDepth) && IsBalance(pRoot->m_pRight,nRightDepth))
	{
		nDiff = nLeftDepth - nRightDepth;
		if (nDiff >= -1 && nDiff <= 1)
		{
			nDepth =  max(nLeftDepth,nRightDepth) + 1;
			return true;
		}
	}
	return false;
}

bool IsBalance(BinaryTreeNode* pRoot)
{
	if (pRoot == NULL)
	{
		cout<<"空树!"<<endl;
		return false;
	}
	int nDepth = 0;
	return IsBalance(pRoot,nDepth);
}

void Create(BinaryTreeNode*& pRoot)    
{    
	char newData;    
	cin >> newData;    
	if ('#' == newData)    
	{    
		pRoot = NULL;    
	}    
	else    
	{    
		pRoot = new BinaryTreeNode;    
		pRoot->m_nValue = newData;    
		Create(pRoot->m_pLeft);    
		Create(pRoot->m_pRight);     
	}    
}    

int main()
{
	
	BinaryTreeNode* pRoot = NULL;
	Create(pRoot);
	if (IsBalance(pRoot))
	{
		cout<<"平衡二叉树!"<<endl;
	}
	else
	{
		cout<<"不是平衡二叉树!"<<endl;
	}
	system("pause");
	return 1;
}

注意：

树的构建类似树的子结构，这里不再详述。