Validate Single Binary Tree

[hamidgasmi]

Given a list of TreeNodes (see the definition below). Find out if all these nodes belong to the same valid binary tree.

Example 1:

Let's say we have the following binary tree
        1
       ↙ ↘
      2   3
         ↙
        4

We can create it like this
TreeNode n1 = new TreeNode(1);
TreeNode n2 = new TreeNode(2);
TreeNode n3 = new TreeNode(3);
TreeNode n4 = new TreeNode(4);

n1.left = n2;
n1.right = n3;
n3.left = n4;

Input: [n4, n2, n3, n1]
Output: true

Example 2:

        1
       ↙  ↘
      2    3
       ↘  ↙
        4

TreeNode n1 = new TreeNode(1);
TreeNode n2 = new TreeNode(2);
TreeNode n3 = new TreeNode(3);
TreeNode n4 = new TreeNode(4);

n1.left = n2;
n1.right = n3;
n2.right = n4;
n3.left = n4;

Input: [n2, n3, n4, n1]
Output: false

Example 3:

	  1
	⤤ ⤦
	 2 

TreeNode n1 = new TreeNode(1);
TreeNode n2 = new TreeNode(2);

n1.left = n2;
n2.left = n1;

Input: [n1, n2]
Output: false

Example 4:

        1             4
       ↙ ↘        ↙  ↘
      2   3      5     6

TreeNode n1 = new TreeNode(1);
TreeNode n2 = new TreeNode(2);
TreeNode n3 = new TreeNode(3);
TreeNode n4 = new TreeNode(4);
TreeNode n5 = new TreeNode(5);
TreeNode n6 = new TreeNode(6);

n1.left = n2;
n1.right = n3;

n4.left = n5;
n4.right = n6;

Input: [n2, n6, n4, n1, n3, n5]
Output: false

Example 5:

        1
       ↙ ↘
      2   3
         ↙
        4

TreeNode n1 = new TreeNode(1);
TreeNode n2 = new TreeNode(2);
TreeNode n3 = new TreeNode(3);
TreeNode n4 = new TreeNode(4);

n1.left = n2;
n1.right = n3;
n3.left = n4;

Input: [n2, n3, n1]
Output: false
Explanation: l4 is a part of the tree but it's missing in the input list so return false.

TreeNode class:

class TreeNode:
   def __init__(self, val = 0, left = None, right: None):
      self.val = val
      self.left = left
      self.right = None

The function:

def is_binary_tree(nodes) -> bool:
   pass

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