-
Notifications
You must be signed in to change notification settings - Fork 0
/
450. Delete Node in a BST(c++)
92 lines (79 loc) · 1.91 KB
/
450. Delete Node in a BST(c++)
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
Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.
Basically, the deletion can be divided into two stages:
Search for a node to remove.
If the node is found, delete the node.
Note: Time complexity should be O(height of tree).
Example:
root = [5,3,6,2,4,null,7]
key = 3
5
/ \
3 6
/ \ \
2 4 7
Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the following BST.
5
/ \
4 6
/ \
2 7
Another valid answer is [5,2,6,null,4,null,7].
5
/ \
2 6
\ \
4 7
--------------------------------------------------xxxxxxxxxxxxxxxxxxxxxxxxxxx------------------------------------------
CODE:-
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int max(TreeNode * root)
{
if(root->right==NULL)
{
return root->val;
}
else
{
return max(root->right);
}
}
TreeNode* deleteNode(TreeNode* root, int key) {
if(root==NULL)
{
return NULL;
}
if(root->val>key)
{
root->left=deleteNode(root->left,key);
}
else if(root->val<key)
{
root->right=deleteNode(root->right,key);
}
else
{
if(root->left==NULL || root->right==NULL)
{
return root->left==NULL?root->right:root->left;
}
else
{
int lmax=max(root->left);
root->val=lmax;
root->left=deleteNode(root->left,lmax);
}
}
return root;
}
};