判斷二差樹是否爲平衡二叉樹

http://blog.csdn.net/zhaojinjia/article/details/11891523

來自劍指offer

求樹的深度

用遞歸做很簡單，只要知道遞歸出口語句的別寫錯。

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);
}

判斷該樹是否爲平衡二叉樹

方法一：調用上述函數求每個節點的左右孩子深度

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：由於上述方法在求該結點的的左右子樹深度時遍歷一遍樹，再次判斷子樹的平衡性時又遍歷一遍樹結構，造成遍歷多次。因此方法二是一邊遍歷樹一邊判斷每個結點是否具有平衡性。

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)  //原文||是錯誤的
        {
            *depth = 1+(nLeftDepth > nRightDepth ? nLeftDepth : nRightDepth);
            return true;
        }
    }

    return false;
}

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

    return IsBalanced(pRoot, &depth);
}