In a binary tree, a lonely node is a node that is the only child of its parent node. The root of the tree is not lonely because it does not have a parent node.
Given the root of a binary tree, return an array containing the values of all lonely nodes in the tree. Return the list in any order.
Example 1:
Input: root = [1,2,3,null,4]
Output: [4]
Explanation: Light blue node is the only lonely node.
Node 1 is the root and is not lonely.
Nodes 2 and 3 have the same parent and are not lonely.
Example 2:
Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2]
Output: [6,2]
Explanation: Light blue nodes are lonely nodes.
Please remember that order doesn't matter, [2,6] is also an acceptable answer.
Example 3:
Input: root = [11,99,88,77,null,null,66,55,null,null,44,33,null,null,22]
Output: [77,55,33,66,44,22]
Explanation: Nodes 99 and 88 share the same parent. Node 11 is the root.
All other nodes are lonely.
Constraints:
The number of nodes in the tree is in the range [1, 1000].
1 <= Node.val <= 106
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:defgetLonelyNodes(self,root:Optional[TreeNode])->List[int]:defdfs(root):ifrootisNoneor(root.leftisNoneandroot.rightisNone):returnifroot.leftisNone:ans.append(root.right.val)ifroot.rightisNone:ans.append(root.left.val)dfs(root.left)dfs(root.right)ans=[]dfs(root)returnans
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcgetLonelyNodes(root*TreeNode)[]int{ans:=[]int{}vardfsfunc(*TreeNode)dfs=func(root*TreeNode){ifroot==nil||(root.Left==nil&&root.Right==nil){return}ifroot.Left==nil{ans=append(ans,root.Right.Val)}ifroot.Right==nil{ans=append(ans,root.Left.Val)}dfs(root.Left)dfs(root.Right)}dfs(root)returnans}