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Copy path1448. Count Good Nodes in Binary Tree.cpp
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1448. Count Good Nodes in Binary Tree.cpp
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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].
Hide Hint #1
Use DFS (Depth First Search) to traverse the tree, and constantly keep track of the current path maximum.
/**
* 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 res = 0;
void traversal(TreeNode *root, int maxi) {
if (!root) return;
if (root->val >= maxi) res++;
maxi = max(maxi, root->val);
traversal(root->left, maxi);
traversal(root->right, maxi);
}
int goodNodes(TreeNode* root) {
if (!root) return res;
traversal(root, root->val);
return res;
}
};