剑指offer——二叉树的深度与平衡二叉树的判断

通过后续遍历，可以减少重复访问

#include <iostream>
#include <string>
using namespace std;

struct BinaryTreeNode
{
    int m_data;
    BinaryTreeNode* m_left;
    BinaryTreeNode* m_right;
};

int TreeDepth(BinaryTreeNode* pRoot)
{
    if (pRoot==NULL)
    {
        return 0;
    }
    int nleft=TreeDepth(pRoot->m_left);
    int nright=TreeDepth(pRoot->m_right);
    return (nleft>nright)?(nleft+1):(nright+1);
}

bool IsBalance(BinaryTreeNode* pRoot,int *pDepth)
{
    if (pRoot==NULL)
    {
        *pDepth=0;
        return true;
    }
    int left,right;
    if(IsBalance(pRoot->m_left,&left)&&IsBalance(pRoot->m_right),&right)
    {
        int diff=left-right;
        if (diff<=1&&diff>=-1)
        {
            *pDepth=1+(left>right)?left:right;
            return true;
        }
    }
    return false;
}