Create 543_diameter_of_binary_tree.md

Author: Joseph Luce

leetcode/easy/543_diameter_of_binary_tree.md

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+# 543. Diameter of Binary Tree
+
+## Best Solution
+- Runtime: O(N)
+- Space: O(N)
+- N = Number of nodes in binary tree
+
+Take this binary tree example:
+```
+    1
+   / \
+  2   3
+ / \
+4   5
+```
+
+In the perspective of node 1, what do you need? 
+- You will need to know what the current longest path found so far from either left or right is.
+- You will also need to know what the longest 'connectable' path is.
+
+Connectable path means taking your child's connectable path and connecting it with yourself, hence, creating a long chain.
+This is completely separate from the curent longest path. 
+You want to know if you can create a longer chain than the current longest path found so far.
+
+```
+class Solution:
+    def diameterOfBinaryTree(self, root: TreeNode) -> int:
+        def length_helper(root):
+            if root is None:
+                return 0, 0
+            # Figure out the longest right and left
+            # Then see if we can use either right or left to create a longer path
+            longest_right, n_right_connectable_nodes = length_helper(root.right)
+            longest_left, n_left_connectable_nodes = length_helper(root.left)
+            n_connected_nodes = n_right_connectable_nodes + n_left_connectable_nodes + 1
+            local_longest = max(longest_right, longest_left, n_connected_nodes)
+            best_n_connectable_nodes = max(n_right_connectable_nodes+1, n_left_connectable_nodes+1)
+            return local_longest, best_n_connectable_nodes
+        if root is None:
+            return 0
+        longest_path, longest_connectable_path = length_helper(root)
+        return longest_path-1
+```