判断树是否为平衡二叉树

来自剑指offer

求树的深度

用递归做很简单，只要知道递归出口语句的别写错。

struct BinaryTreeNode
{
	int m_Value;
	BinaryTreeNode* m_pLeft;
	BinaryTreeNode* m_pRight;
};

int TreeDepth(BinaryTreeNode* pRoot)
{
	if (pRoot == NULL)
		return 0;

	int nLeftDepth = TreeDepth(pRoot->m_pLeft);
	int nRightDepth = TreeDepth(pRoot->m_pRight);

	return (nLeftDepth>nRightDepth)?(nLeftDepth+1):(nRightDepth+1);
}

判断该树是否为平衡二叉树

方法一：调用上述函数求每个节点的左右孩子深度

bool IsBalanced(BinaryTreeNode* pRoot)
{
	if(pRoot== NULL)
		return true;

	int nLeftDepth = TreeDepth(pRoot->m_pLeft);
	int nRightDepth = TreeDepth(pRoot->m_pRight);
	int diff = nRightDepth-nLeftDepth;

	if (diff>1 || diff<-1)
		return false;

	return IsBalanced(pRoot->m_pLeft)&&IsBalanced(pRoot->m_pRight);
}

方法二：由于上述方法在求该结点的的左右子树深度时遍历一遍树，再次判断子树的平衡性时又遍历一遍树结构，造成遍历多次。因此方法二是一边遍历树一边判断每个结点是否具有平衡性。

bool IsBalanced(BinaryTreeNode* pRoot, int* depth)
{
	if(pRoot== NULL)
	{
		*depth = 0;
		return true;
	}

	int nLeftDepth,nRightDepth;
	bool bLeft= IsBalanced(pRoot->m_pLeft, &nLeftDepth);
	bool bRight = IsBalanced(pRoot->m_pRight, &nRightDepth);
	
	if (bLeft && bRight)
	{
		int diff = nRightDepth-nLeftDepth;
		if (diff<=1 || diff>=-1)
		{
			*depth = 1+(nLeftDepth > nRightDepth ? nLeftDepth : nRightDepth);
			return true;
		}
	}
	
	return false;
}

bool IsBalanced(BinaryTreeNode* pRoot)
{
	int depth = 0;

	return IsBalanced(pRoot, &depth);
}