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Inorder Successor in BST II
Bookmark
Find Root + Inorder
Parent Pointers
Input
Has right subtree (successor = 6)
Walk up to parent (successor = 5)
No successor (rightmost node)
Custom
tree
=
[5, 3, 7, 2, 4, 6, 8]
,
nodeValue
=
5
node = 5
Step 1: climb to the root
5
3
7
2
4
6
8
root
inorder
empty
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 1: climb to the root
5
3
7
2
4
6
8
root
inorder
empty
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 2: collect the inorder sequence
5
3
7
2
4
6
8
current
inorder
empty
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 2: collect the inorder sequence
5
3
7
2
4
6
8
current
inorder
2
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 2: collect the inorder sequence
5
3
7
2
4
6
8
current
inorder
2
3
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 2: collect the inorder sequence
5
3
7
2
4
6
8
current
inorder
2
3
4
5
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 2: collect the inorder sequence
5
3
7
2
4
6
8
current
inorder
2
3
4
5
6
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 2: collect the inorder sequence
5
3
7
2
4
6
8
current
inorder
2
3
4
5
6
7
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 2: collect the inorder sequence
5
3
7
2
4
6
8
current
inorder
2
3
4
5
6
7
8
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 3: scan for the given node
5
3
7
2
4
6
8
current
inorder
2
3
4
5
6
7
8
Successor: null
Current
Visited
Given node
Successor
node = 5
Step 3: scan for the given node
5
3
7
2
4
6
8
successor
inorder
2
3
4
5
6
7
8
Successor: 6
Current
Visited
Given node
Successor
algo
master
.
io
Step:
root = node (5)
0 / 41
Input
Has right subtree (successor = 6)
Walk up to parent (successor = 5)
No successor (rightmost node)
Custom
tree
=
[5, 3, 7, 2, 4, 6, 8]
,
nodeValue
=
5
0 / 41
algo
master
.
io
Step:
root = node (5)
