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
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# 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 = rightclassSolution:defgoodNodes(self,root:TreeNode)->int:defdfs(root:TreeNode,mx:int):ifrootisNone:returnnonlocalansifmx<=root.val:ans+=1mx=root.valdfs(root.left,mx)dfs(root.right,mx)ans=0dfs(root,-1000000)returnans
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcgoodNodes(root*TreeNode)(ansint){vardfsfunc(*TreeNode,int)dfs=func(root*TreeNode,mxint){ifroot==nil{return}ifmx<=root.Val{ans++mx=root.Val}dfs(root.Left,mx)dfs(root.Right,mx)}dfs(root,-10001)return}