A: 二叉查找树由节点组成,所以我们要定义的第一个对象就是Node
B: 没有任何作用
C: 表示树的路径
D: 以上都不对
举一反三
- voidABC(BTNode*BT){ifBT{ABC(BT->left);ABC(BT->right);cout<data<
- Eat to your ______,and drink to your_______.? right, left|right, right;|left, left|left, right
- In left sideslip, the vertical tail will generate a side force and yawing moment. A: left, left B: left, right C: right, left D: right, right
- 设\( A \)为\( n \) 阶方阵, \( B \)是\( A \)经过若干次初等变换后得到的矩阵,则( ) A: \( \left| A \right| = \left| B \right| \) B: \( \left| A \right| \ne \left| B \right| \) C: 若\( \left| A \right| = 0 \) ,则必有 \( \left| B \right| = 0 \) D: 若\( \left| A \right| > 0 \),则一定有\( \left| B \right| > 0 \)
- 设\( A,B \)均为\( n \)阶方阵,则必有( ) A: \( \left| {A + B} \right| = \left| A \right| + \left| B \right| \) B: \( AB = BA \) C: \( \left| {AB} \right| = \left| {BA} \right| \) D: \( {\left( {A + B} \right)^{ - 1}} = {A^{ - 1}} + {B^{ - 1}} \)
内容
- 0
若\({y_1}\left( x \right), {y_2}\left( x \right)\)都是\(y' + P\left( x \right)y = Q\left( x \right)\)的特解,且 \({y_1}\left( x \right), {y_2}\left( x \right)\) 线性无关,则通解可表为\(y\left( x \right) = {y_1}\left( x \right) + C\left[ { { y_1}\left( x \right) - {y_2}\left( x \right)} \right]\)。
- 1
在网页中插入两个id名分别为left和right的标签,()两个元素都左浮动。 A: left, B: right{float:left;} C: left, D: right{float:right;} E: left{float:left;} F: right{float:right;} G: left{width:100px;float:left;} H: right{margin-left:100px;}
- 2
在网页中插入两个id名分别为left和right的标签,()两个元素都右浮动。 A: left, B: right{float:left;} C: left, D: right{float:right;} E: left{float:left;} F: right{float:right;} G: left{width:100px;float:left;} H: right{margin-left:100px;}
- 3
设\( A,\;B \) 均为\( n \) 阶方阵,则必有( ). A: \( {(A + B)^2} = {A^2} + 2AB + {B^2} \) B: \( \left| {A + B} \right| = \left| A \right| + \left| B \right| \) C: \( \left| {AB} \right| = \left| A \right|{\kern 1pt} \left| B \right| \) D: \( {\left( {AB} \right)^{\rm T}} = {A^{\rm T}}{B^{\rm T}} \)
- 4
设\( A,B \) 为方阵,则 \( \left| {AB} \right| = \,\left| A \right|\,\left| B \right| \)。( )