110. 平衡二叉树

一、递归

单次递归时，判断左右子树是否是平衡树，若不是，返回-1，若是，但输得出高度

class Solution {
public:
    // 返回以该节点为根节点的二叉树的高度，如果不是二叉搜索树了则返回-1
    int getHeight(TreeNode* node) {
        if (node == NULL) {
            return 0;
        }
        int leftHeight = getHeight(node->left);
        if (leftHeight == -1) return -1;
        int rightHeight = getHeight(node->right);
        if (rightHeight == -1) return -1;
        return abs(leftHeight - rightHeight) > 1 ? -1 : 1 + max(leftHeight, rightHeight);
    }
    bool isBalanced(TreeNode* root) {
        return getHeight(root) == -1 ? false : true;
    }
};