《剑指offer》:[39-1]判断是否为平衡二叉树

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题目：输入一棵二叉树的结点，判断该树是不是平衡二叉树。如果某二叉树中任意结点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。

#include <iostream>
using namespace std;
struct BinaryTree
{
	int data;
	BinaryTree *pLeft;
	BinaryTree *pRight;
};
BinaryTree *pRoot1=NULL;
int arr[7]={8,5,15,3,7,16,6};
void InserTree(BinaryTree **root,int data)
{
	BinaryTree *temp=new BinaryTree;
	temp->data=data;
	temp->pLeft=temp->pRight=NULL;
	if(NULL==*root)
	{
		*root=temp;
	}
	else
	{
		BinaryTree *current=*root;
		BinaryTree *back=NULL;
		while(current)
		{
			back=current;
			if(data > current->data)
				current=current->pRight;
			else
				current=current->pLeft;
		}
		if(NULL==back->pLeft)
			back->pLeft=temp;
		else
			back->pRight=temp;
	}
}
void CreateTreeLoop(BinaryTree **root,int *array,int length)
{
	for(int i=0;i<length;i++)
		InserTree(root,array[i]);
}
void Inorder(BinaryTree *root)
{
	if(NULL!=root)
	{
		Inorder(root->pLeft);
		cout<<root->data<<" ";
		Inorder(root->pRight);
	}
}
int TreeDepth(BinaryTree*root)
{
	if(NULL==root)
		return 0;
	int left=TreeDepth(root->pLeft);
	int right=TreeDepth(root->pRight);
	return left>right?(left+1):(right+1);
}
bool IsBalance(BinaryTree *root)
{
	if(NULL==root)
		return true;
	int leftIsBla=TreeDepth(root->pLeft);
	int rightIsBla=TreeDepth(root->pRight);
	int diff=leftIsBla-rightIsBla;
	if(diff>1 || diff<-1  )
		return false;
	return IsBalance(root->pLeft) && IsBalance(root->pRight);
}


int main()
{
	CreateTreeLoop(&pRoot1,arr,7);
	cout<<"中序遍历：";
	Inorder(pRoot1);
	cout<<endl;
	bool result=IsBalanced(pRoot1);
	if(result)
		cout<<"The tree is balance!"<<endl;
	else
		cout<<"The tree is not a balance tree!"<<endl;
	system("pause");
	return 0;
}

运行结果：

bool IsBalanceHelp(BinaryTree *root,int *depth)
{
	if(root==NULL)
	{
		*depth=0;
		return true;
	}
	int left,right;
	if(IsBalanceHelp(root->pLeft,&left)&& IsBalanceHelp(root->pRight,&right))
	{
		int diff=left-right;
		if(diff<=1 && diff>=-1)
		{
			*depth=1+(left>right?left:right); //当遍历每个结点时，记录下该结点的深度，下次就不用再遍历已经遍历过的结点；
			return true;
		}
	}
	return false;
}
bool IsBalanced(BinaryTree *root)
{
	int depth=0;
	return IsBalanceHelp(root,&depth);
}

《剑指offer》:[39-1]判断是否为平衡二叉树

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原文地址：http://blog.csdn.net/gogokongyin/article/details/51712659