题目：
输入一颗二叉树的根节点，求该树的深度。从根结点到叶结点依次经过的结点（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。

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

解法：

int treeDepth(BinaryTreeNode* pRoot) {
    if (pRoot == 0)    return 0;
    int left    =  treeDepth(pRoot->m_pLeft);
    int right  =  treeDepth(pRoot->m_pRight);
    return 1+ (left > right ? left : right); 
}

拓展
判断一颗二叉树是否是一颗平衡二叉树。平衡二叉树，对任何一个结点，左右子树的深度差距不超过1.
解法一：
从根节点开始，对每一个结点判断左右子树是否平衡。判断的方法：求左右子树的深度，如果深度相差不超过1，即可判断为平衡二叉树

    bool isBalanced(BinaryTreeNode* pRoot) {
        if (pRoot == 0)    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(pRoot->m_pLeft) && isBalanced(pRoot->m_pRight);
    }

该方法，会重复遍历结点

解法二：

    bool isBalanced(BinaryTreeNode* pRoot, int *depth) {
        if (pRoot == 0) {
            *depth = 0;
            return true;
        }
        int left = 0;
        int right = 0;
        bool leftBalanced = isBalanced(pRoot->m_pLeft, &left);
        bool rightBalanced = isBalanced(pRoot->m_pRight, &right);
        int diff = left - right;
        if (leftBalanced && rightBalanced && (diff <= 1 && diff >= -1)) {
            *depth = 1 + (left > right ? left : right);
            return true;
        }
        return false;
    }