判断一个二叉树是否是平衡二叉树

题目：判断一个二叉排序树是否是平衡二叉树

思路:利用递归判断左右子树的深度是否相差1来判断是否是平衡二叉树。

#include<stdio.h>
#include "stdafx.h"

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

BinaryTreeNode* CreateBinaryTreeNode(int value)
{
    BinaryTreeNode* pNode = new BinaryTreeNode();
    pNode->m_nValue = value;
    pNode->m_pLeft = NULL;
    pNode->m_pRight = NULL;
}

void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
    if(pParent != NULL)
    {
        pParent->m_pLeft = pLeft;
        pParent->m_pRight = pRight;
    }
}

void PrintTreeNode(BinaryTreeNode* pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->m_nValue);
       
        if(pNode->m_pLeft != NULL)
            printf("value of its left child is: %d.\n", pNode->m_pLeft->m_nValue);
        else
            printf("left child is null.\n");
       
        if(pNode->m_pRight != NULL)
            printf("value of its right child is: %d.\n",pNode->m_pRight->m_nValue);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }
    printf("\n");
}

void PrintTree(BinaryTreeNode* pRoot)
{
    PrintTreeNode(pRoot);
   
    if(pRoot != NULL)
    {
        if(pRoot->m_pLeft != NULL)
            PrintTree(pRoot->m_pLeft);
       
        if(pRoot->m_pRight != NULL)
            PrintTree(pRoot->m_pRight);
    }
}

void DestroyTree(BinaryTreeNode* pRoot)
{
    if(pRoot != NULL)
    {
        BinaryTreeNode* pLeft = pRoot->m_pLeft;
        BinaryTreeNode* pRight = pRoot->m_pRight;
       
        delete pRoot;
        pRoot = NULL;
       
        DestroyTree(pLeft);
        DestroyTree(pRight);
    }
}

//========================方法1==============================
int TreeDepth(BinaryTreeNode* pRoot)
{
    if(pRoot == NULL)
        return 0;
   
    int nLeft = TreeDepth(pRoot->m_pLeft);
    int nRight = TreeDepth(pRoot->m_pRight);
   
    return (nLeft > nRight) ? (nLeft + 1) : (nRight + 1);
}

bool IsBalanced_Solution1(BinaryTreeNode* pRoot)
{
    if(pRoot == NULL)
        return true;
   
    int left = TreeDepth(pRoot->m_pLeft);
    int right = TreeDepth(pRoot->m_pRight);
    int diff = left - right;
    if(diff > 1 || diff < -1)
        return false;
   
    return IsBalanced_Solution1(pRoot->m_pLeft)
        && IsBalanced_Solution1(pRoot->m_pRight);
}

//=====================方法2===========================
bool IsBalanced(BinaryTreeNode* pRoot, int* pDepth);

bool IsBalanced_Solution2(BinaryTreeNode* pRoot)
{
    int depth = 0;
    return IsBalanced(pRoot, &depth);
}

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;
}