方法一：

//求當(dāng)前節(jié)點(diǎn)的高度

int level(TreeNode *root)

{

    if(root == NULL) return 0;

    return 1 + max(level(root->left),level(root->right));

}

bool isBalanced(TreeNode *root)

{

    if(root == NULL)    return true;

    int factor = abs(level(root->left) - level(root->right));

    return factor < 2 && isBalanced(root->left) && isBalanced(root->right);

}

方法二：一種更好的方法

isBalanced返回當(dāng)前節(jié)點(diǎn)的高度,，-1代表樹不是平衡二叉樹，將計(jì)算結(jié)果自底向上傳遞,，并且確保每個(gè)節(jié)點(diǎn)只計(jì)算一次,，復(fù)雜度為O(n)。

int isBalancedHelper(TreeNode *root)

{

    if(root == NULL) return 0;

    int Lh = isBalancedHelper(root->left);

    if(Lh == -1)     return -1;

    int Rh = isBalancedHelper(root->right);

    if(Rh == -1)    return -1;

    if(abs(Lh - Rh) > 1)    return -1;

    return max(Lh,Rh) + 1;

}

bool isBalancedTree(TreeNode *root)

{

    if(root == NULL)    return true;

    return (isBalancedHelper(root) != -1);

}