我的树的平衡部分遇到了问题。我在递归插入后调用了 checkBal。如果我尝试添加 5、2 和 4,它会检查 2 的平衡并继续回到 5,然后进入右旋转的 rotateLeft 部分,这是正确的。但是第二行的 rotateLeft 函数出错。
这个实现有什么问题?我到处搜索并将我所做的与人们谈论它是如何完成的方式进行比较。我终于让一切正常了。最后我忘记将 N 设置为 K 了。
//==============================================================================
//===== Set Balance ============================================================
sNode<T>* checkBal(sNode<T> *locRoot)
{
// Make sure to check the children are balanced as well.
if (locRoot->left != NULL)
locRoot->left = checkBal(locRoot->left);
if (locRoot->right != NULL)
locRoot->right = checkBal(locRoot->right);
if(getHeight(locRoot->left) - getHeight(locRoot->right) > 1)
{
if(getHeight(locRoot->left->right) > getHeight(locRoot->left->left))
locRoot->left = rotateRight(locRoot->left);
locRoot = rotateLeft(locRoot);
}
else if(getHeight(locRoot->right) - getHeight(locRoot->left) > 1)
{
if(getHeight(locRoot->right->left) > getHeight(locRoot->right->right))
locRoot->right = rotateLeft(locRoot->right);
locRoot = rotateRight(locRoot);
}
updateHeights(locRoot);
return locRoot;
}
/*
Extream cases of balancing a tree requires a double rotation
A
\
D
/
B
'A' is the current root
If right->left (grandchild) is larger then the right->right (grandchild)
First Right rotate the child then left rotate the parent
left > right by 2 or more
left.left < left.right (Double Right Rotation)
left.left >= left.right (Single Right Rotation)
right > left by 2 or more
right.right < right.left (Double Left Rotation)
right.right >= right.left (Single Left Rotation)
*/
sNode<T>* rotateRight(sNode<T> *N) const
{
/*
N K
/ \ / \
(a) K => N (c)
/ \ / \
(b) (c) (a) (b)
*/
// K is going to be our new Parent
// Move (c) from K->right to N->left
// Set K->right to be N
// Return the new parent node to update the one above.
sNode<T> *K = N->right;
N->right = K->left;
K->left = N;
return N = K;
}