剑指offer---平衡二叉树

方法一：基本方法 不做解释了

class Solution 
{
public:
    int shendu(TreeNode* &root)
    {
        if(root==NULL)
        {
            return 0;
        }
        
        int zuozishu=shendu(root->left);
        int youzishu=shendu(root->right);
        
        return (zuozishu<youzishu?youzishu+1:zuozishu+1);
    }
    
    //遍历
    bool IsBalanced_Solution(TreeNode* pRoot)
    {
        if(pRoot==NULL)
        {
            return true;
        }
        
        int leftdepth=shendu(pRoot->left);
        int rightdepth=shendu(pRoot->right);
        int chazhi=abs(leftdepth-rightdepth);
        
        if(chazhi>1)
        {
            return false;
        }
        
        return(IsBalanced_Solution(pRoot->right)&&IsBalanced_Solution(pRoot->left));
    }
};

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);  
}