推断二叉树是不平衡树 代码(C)

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题目: 输入一颗二叉树的根结点, 推断该树是不是平衡二叉树.

二叉平衡树: 随意结点的左右子树的深度相差不超过1.

使用后序遍历的方式, 而且保存左右子树的深度, 进行比較.

代码:

/* * main.cpp * *  Created on: 2014.6.12 *      Author: Spike *//*eclipse cdt, gcc 4.8.1*/#include 
     
      #include 
      
       #include 
       
        struct BinaryTreeNode {	int m_nValue;	BinaryTreeNode* m_pLeft;	BinaryTreeNode* m_pRight;};bool IsBalanced(BinaryTreeNode* pRoot, int* pDepth) {	if (pRoot == NULL) {		*pDepth = 0;		return true;	}	int left, right;	if (IsBalanced(pRoot->m_pLeft, &left) && IsBalanced(pRoot->m_pRight, &right)) {		int diff = left - right;		if (diff>=-1 && diff<=1) {			*pDepth = 1 + (left>right?left:right);			return true;		}	}	return false;}bool IsBalanced(BinaryTreeNode* pRoot) {	int depth = 0;	return IsBalanced(pRoot, &depth);}BinaryTreeNode* init(void) {	BinaryTreeNode* pRoot = new BinaryTreeNode(); pRoot->m_nValue = 1;	BinaryTreeNode* pNode2 = new BinaryTreeNode(); pNode2->m_nValue = 2;	BinaryTreeNode* pNode3 = new BinaryTreeNode(); pNode3->m_nValue = 3;	BinaryTreeNode* pNode4 = new BinaryTreeNode(); pNode4->m_nValue = 4;	BinaryTreeNode* pNode5 = new BinaryTreeNode(); pNode5->m_nValue = 5;	BinaryTreeNode* pNode6 = new BinaryTreeNode(); pNode6->m_nValue = 6;	BinaryTreeNode* pNode7 = new BinaryTreeNode(); pNode7->m_nValue = 7;	pRoot->m_pLeft = pNode2; pRoot->m_pRight = pNode3;	pNode2->m_pLeft = pNode4; pNode2->m_pRight = pNode5;	pNode4->m_pLeft = NULL; pNode4->m_pRight = NULL;	pNode5->m_pLeft = pNode7; pNode5->m_pRight = NULL;	pNode7->m_pLeft = NULL; pNode7->m_pRight = NULL;	pNode3->m_pLeft = NULL; pNode3->m_pRight = pNode6;	pNode6->m_pLeft = NULL; pNode6->m_pRight = NULL;	return pRoot;}int main(void){	BinaryTreeNode* pRoot = init();	bool result = IsBalanced(pRoot);	printf("result = %s\n", result==false?"false":"true");    return 0;}
       
      
     

输出:

result = true