1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
|
[\<- 05/11](05-11.md)
---
## BST - Insertion
- To insert data, we need to follow the branches to an empty subtree and then insert the new node
- All inserts take place at a leaf or at a leaflike node - a node that has only one subtree
- Example: insert 15, 13
![example tree](05-13_img1.png)
- For 15
- 15 is less than 17
- 15 is greater than 6
- 15 is greater than 14 -> insert in the right child of the 14 node
- For 13
- 13 is less than 17
- 13 is greater than 6
- 13 is less than 14 -> insert in the left child of the 14 node
## Exercise
- Give me an arbitrary number and let's insert it into the tree
- 50
- 30
- 60
- 20
- 100
- 10
- for 50
- We start with an empty tree, so 50 becomes the root
- for 30
- 30 is less than 50 -> insert in the left child of the root
- for 60
- 60 is greater than 50 -> insert in the right child of the root
- for 20
- 20 is less than 50
- 20 is less than 30 -> insert in the left child of the 30 node
- for 100
- 100 is greater than 50
- 100 is greater than 60 -> insert in the right child of the 60 node
- for 10
- 10 is less than 50
- 10 is less than 30
- 10 is less than 20 -> insert in the left child of the 20 node
- What's the runtime for something like this? **O(h)** (h being the height of the tree)
## Insertion Code
```
bool insertNode(NODE *root, NODE *np){
assert(np != NULL);
//Base case
if(root == NULL) root = np;
//If the node to be inserted is smaller than the root's data, it lies in left subtree
if(np->data < root->data) return insertNode(root->left, np);
//If the node to be inserted is greater than the root's data, it lies in the right subtree
else if(np->data > root->data) return insertNode(root->right, np);
//if x is the same as root's data
else return false;
}
```
- This code doesn't work (why?)
- It creates a disconnected tree
- The solution? Double pointers!
- What is the correct code?
```
bool insertNode(NODE **root, NODE *np){
assert(np != NULL);
//Base case
if(*root == NULL) *root = np;
//If the node to be inserted is smaller than the root's data, it lies in left subtree
if(np->data < (*root)->data) return insertNode(&(root)->left, np);
//If the node to be inserted is greater than the root's data, it lies in the right subtree
else if(np->data > (*root)->data) return insertNode(&(*root)->right, np);
//if x is the same as root's data
else return false;
}
```
- Notice that in the above version, root is a double pointer
There is another way!
```
NODE *insertNode(NODE *root, NODE *np){
assert(np != NULL);
//Base case
if(root == NULL) return np;
//If the node to be inserted is smaller than the root's data, it lies in left subtree
if(np->data < root->data) root->left = insertNode(root->left, np);
//If the node to be inserted is greater than the root's data, it lies in the right subtree
else if(np->data > root->data) root->right = insertNode(root->right, np);
//if x is the same as root's data
else return false;
}
```
- Notice that the above code returns a NODE * (and not a boolean)
- Keep in mind that the above code is relatively complex
## BST - Deletion
- Deleting nodes with no children is easy, you just search for and free them (doesn't affect the rest of the tree)
- Deleting nodes with one child is a little challenging, because we want to maintain the node's child, but also to remove that node
- take the "abandoned" child and have the "grandparent" node take care of it
- Deleting a node with two children is more challenging
- The "grandparent" node only has one available pointer, but now there are two "abandoned" children
- The "abandoned" children must form their own BST, the root of this new tree will be pointed to by the "grandparent"
### Turning two subtrees into a tree
- Either take the minimum value from the right or the maxiumum value from the left
- They will become the root, basically replacing the removed parent
### Summary
- Assume the node to be deleted is `Ndel`, there are four cases
- `Ndel` has **no children** -> delete it
- `Ndel` has **only a right subtree** -> delete `Ndel` and attach its right subtree to `Ndel`'s parent
- `Ndel` has **only a left subtree** -> delete `Ndel` and attach the left subtree to `Ndel`'s parent
- `Ndel` has **two subtrees** -> replace `Ndel`'s data by either the largest node in its left subtree or the smalles node in its right subtree
## BST - Deletion Big-O
- What's the worst-case big O?
- The only looping in the process is the searching -> O(h)
## Let's Work on the Code
- One Challenge
- Deletion may change the structure of the subtree, leading to a different root of the subtree. How could we connect the parent with the new root of the subtree?
- Solution: We pass the new root of the subtree (after deletion) back to the parent
### Deletion Code
```
NODE *deleteNODE(NODE *root, int x, bool *found){
//base case
if(root == NULL){
*found = false;
return root;
}
//If the node to be deleted is smaller than the root's data, it lies in left subtree
if(x < root->data) root->left = deleteNode(root->left, x, found);
//If the node to be deleted is greater than the root's data, it lies in right subtree
else if(x > root->data) root->right = deleteNode(root->right, x, found);
//if x is same as root's data, this is the node to be deleted
else{
*found = true;
//node with only one child or no child
if(root->left == NULL){
NODE *temp = root->right;
free(root);
return temp;
}
else if(root->right == NULL){
NODE *temp = root->left;
free(root);
return temp;
}
//node with two children: Get the smallest in the right subtree
NODE *temp = minimum(root->right);
//Copy the inorder successor's content to this node
root->data = temp->data;
//Delete the inorder successor
root->right = deleteNode(root->right, temp->data, found);
}
return root;
}
```
---
|SET |Unsorted Array|Sorted Array|Hash Table|Unsorted Linked List|Sorted Linked List |BST |
|---------------|--------------|------------|----------|--------------------|-----------------------------------|-----------------------|
|**Search/Find**|O(n) |O(log(n)) |O(n) |O(n) |O(n) |O(h) (log(n) <= h <= n)|
|**Add** |O(n) |O(n) |O(n) |O(n) |O(n) |O(h) (log(n) <= h <= n)|
|**Remove** |O(n) |O(n) |O(n) |O(n) |O(n) |O(h) (log(n) <= h <= n)|
|**Min/Max** |O(n) |O(1) |O(m) |O(n) |O(1) (assuming fast access to tail)|O(h) (log(n) <= h <= n)|
---
[05/15 ->](05-15.md)
|