Description#
Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.
Return the number of good nodes in the binary tree.
Example 1:
Input: root = [3,1,4,3,null,1,5]
Output: 4
Explanation: Nodes in blue are good.
Root Node (3) is always a good node.
Node 4 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path
Node 3 -> (3,1,3) is the maximum value in the path.
Example 2:
Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.
Example 3:
Input: root = [1]
Output: 1
Explanation: Root is considered as good.
Constraints:
- The number of nodes in the binary tree is in the range
[1, 10^5]. - Each node's value is between
[-10^4, 10^4].
Solutions#
Solution 1#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def goodNodes(self, root: TreeNode) -> int:
def dfs(root: TreeNode, mx: int):
if root is None:
return
nonlocal ans
if mx <= root.val:
ans += 1
mx = root.val
dfs(root.left, mx)
dfs(root.right, mx)
ans = 0
dfs(root, -1000000)
return ans
|
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
| /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int ans = 0;
public int goodNodes(TreeNode root) {
dfs(root, -100000);
return ans;
}
private void dfs(TreeNode root, int mx) {
if (root == null) {
return;
}
if (mx <= root.val) {
++ans;
mx = root.val;
}
dfs(root.left, mx);
dfs(root.right, mx);
}
}
|
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
| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int goodNodes(TreeNode* root) {
int ans = 0;
function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int mx) {
if (!root) {
return;
}
if (mx <= root->val) {
++ans;
mx = root->val;
}
dfs(root->left, mx);
dfs(root->right, mx);
};
dfs(root, -1e6);
return ans;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func goodNodes(root *TreeNode) (ans int) {
var dfs func(*TreeNode, int)
dfs = func(root *TreeNode, mx int) {
if root == nil {
return
}
if mx <= root.Val {
ans++
mx = root.Val
}
dfs(root.Left, mx)
dfs(root.Right, mx)
}
dfs(root, -10001)
return
}
|
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
| /**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function goodNodes(root: TreeNode | null): number {
let ans = 0;
const dfs = (root: TreeNode | null, mx: number) => {
if (!root) {
return;
}
if (mx <= root.val) {
++ans;
mx = root.val;
}
dfs(root.left, mx);
dfs(root.right, mx);
};
dfs(root, -1e6);
return ans;
}
|
