刷代码随想录有感（46）：平衡二叉树

这道题给我的教训：别轻视递归。

题干：

代码：

int geiHeight(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;
 
    if(abs(leftHeight - rightHeight) > 1) return -1;
    
    return 1 + max(leftHeight, rightHeight);
}
 
bool isBalanced(TreeNode* root){
    return getHeight(root) == -1 ? false : true;
}