510. Inorder Successor in BST II #181
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altay9
reviewed
May 8, 2021
| } | ||
| return null; | ||
| } | ||
| return findSuc(start); |
Collaborator
There was a problem hiding this comment.
Can we alter with this:
return findSuc(start.right);
altay9
reviewed
May 8, 2021
Comment on lines
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| while (start.parent != null) { | ||
| if (start.parent.left == start) { | ||
| return start.parent; | ||
| } | ||
| start = start.parent; | ||
| } | ||
| return null; |
Collaborator
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We can combine if and while like this:
public Node inorderSuccessor(Node start) {
if (start == null) {
return null;
}
if (start.right != null) return findSuc(start.right);
while (start.parent != null && start.parent.left != start) {
start = start.parent;
}
return start.parent;
}
Collaborator
Author
|
Would an iterative solution be faster? Such as this. Again, theoretically, it shouldn't be. |
Collaborator
The iterative one is slower. |
altay9
approved these changes
May 8, 2021
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Issue.
We have previously encountered in this question, so it might be helpful to revise that before solving this.
is: inorder successorThe idea is simple:
If a node has a right node, then the
isshould be the leftmost node at that side of tree.Else, if a node has a parent node and its left node is the node, then the
isis the node's parent.Else, if a node has a parent node and its right node is the node, then the
isis the parent's parent if its parent is its corresponding left node, if its however the right node, then we must recursively climb up the tree, until we find a node that is its parent's left node.If all these do not work, then the result is
null, as there is nois.