【剑指offer】推断二叉树平衡

转载请注明出处：http://blog.csdn.net/ns_code/article/details/27242575

    题目：输入一个二叉树的根节点，推断该树是不是平衡二叉树。

假设某二叉树中随意节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。

    剑指offer上给的另外一种思路，用后序遍历真的是将递归发挥的淋漓尽致。先遍历节点的左右子树。左右子树都平衡才来推断该节点是否平衡。假设左右子树中有不平衡的，则直接返回false，避免了从上往下逐个节点地计算深度带来的反复遍历节点。

    代码例如以下：

#include<stdio.h>
#include<stdlib.h>

typedef struct BTNode
{
	char data;
	struct BTNode *pLchild;
	struct BTNode *pRchild;
}BTNode, *BTree;

BTree create_tree();
bool IsBalanced(BTree,int *);
bool IsBalanced(BTree);

int main()
{
	BTree pTree = create_tree();
	if(IsBalanced(pTree))
		printf("Balanced
");
	else
		printf("Not Balanced
");

	return 0;
}

BTree create_tree()
{
	BTree pA = (BTree)malloc(sizeof(BTNode));
	BTree pB = (BTree)malloc(sizeof(BTNode));
	BTree pD = (BTree)malloc(sizeof(BTNode));
	BTree pE = (BTree)malloc(sizeof(BTNode));
	BTree pC = (BTree)malloc(sizeof(BTNode));
	BTree pF = (BTree)malloc(sizeof(BTNode));


	pA->data = 'A';
	pB->data = 'B';
	pD->data = 'D';
	pE->data = 'E';
	pC->data = 'C';
	pF->data = 'F';
	
	pA->pLchild = pB;
	pA->pRchild = pC;
	pB->pLchild = pD;
	pB->pRchild = pE;
	pD->pLchild = NULL;
	pD->pRchild = NULL;
	pE->pLchild = pE->pRchild = NULL;
	pC->pLchild = NULL;
	pC->pRchild = pF;
	pF->pLchild = pF->pRchild = NULL;	
	
	return pA;
}


/*
兴许递归遍历推断二叉树是否平衡
*/
bool IsBalanced(BTree pTree,int *depth)
{
	if(pTree == NULL)
	{
		*depth = 0;
		return true;
	}

	int leftDepth,rightDepth;
	if(IsBalanced(pTree->pLchild,&leftDepth) && IsBalanced(pTree->pRchild,&rightDepth))
	{
		int diff = leftDepth-rightDepth;
		if(diff<=1 && diff>=-1)
		{
			*depth = (leftDepth>rightDepth ? leftDepth:rightDepth) + 1;
			return true;
		}
	}
	return false;
}

/*
封装起来
*/
bool IsBalanced(BTree pTree)
{
	int depth = 0;
	return IsBalanced(pTree,&depth);
}

    測试结果：