题解 | #判断是不是平衡二叉树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param pRoot TreeNode类 
 * @return bool布尔型
 */
//求树高
int tree_deep(struct TreeNode* pRoot){
    if(pRoot == NULL) return 0;
    return tree_deep(pRoot->left)>tree_deep(pRoot->right)?tree_deep(pRoot->left)+1:tree_deep(pRoot->right)+1;
}

//双递归判断是否为平衡二叉
bool IsBalanced_Solution(struct TreeNode* pRoot ) {
    // write code here
    if(pRoot == NULL) return true;
    //左树的高
    int left = tree_deep(pRoot->left);
    //右树的高
    int right = tree_deep(pRoot->right);
    //差在1内
    if(left-right>1||left-right<-1) return false;
    return IsBalanced_Solution(pRoot->left)&&IsBalanced_Solution(pRoot->right);
}