add diagram, 🎶,clean up code 👍

binary-tree-max-diameter-height.py

@@ -1,26 +1,20 @@
-'''
-Question: Find max diameter (width) and height of a binary tree
-
-Height is the number of nodes along the longest path from the node down 
-to the farthest leaf node.
-
-Diameter of a tree (sometimes called the width) is the number of nodes on the 
-longest path between two leaves in the tree. 
-
+#Question: Find max diameter (width, longest path) and height of a binary tree
 
+'''
+NOTES:
 Diameter of a binary tree T will be the largest of the following quantities:
 1. Diameter of T’s left subtree
 2. Diameter of T’s right subtree
 3. Longest path between leaves that goes through the root of T 
-   (this can be computed from the heights of the subtrees of T)
+    - height of left subtrees + height of right subtree + 1
 '''
 
 import sys
 sys.path.append("./mylib")
 import Tree
 
-#Reference: http://tech-queries.blogspot.com/2010/09/diameter-of-tree-in-on.html
 #Solution 1:
+#Reference: http://tech-queries.blogspot.com/2010/09/diameter-of-tree-in-on.html
 #Time complexity: O(n^2) . height() is linear however it is called for every node that results in a complexity of O(n^2). 
 #Time complexity can be reduced to O(n) if we calculate the height with diameter for each node (solution 2)
 def diameter(root):
@@ -40,34 +34,70 @@ def height(root):
 	z = max(x,y) + 1
 	return z
 
+#Solution 2: Optimized to find max diameter and height
 #Reference 1: http://www.careercup.com/question?id=1767700
 #Reference 2: https://crackinterviewtoday.wordpress.com/2010/03/11/diameter-of-a-binary-tree/
-#Solution 2: Optimized to find max diameter and height
 #Time complexity: O(n)
 def diameter_height(tree):	
 	if tree is None: 
 		return (0,0)
 	ld, lh = diameter_height(tree.left)   #Diameter of left subtree
 	rd, rh = diameter_height(tree.right)  #Diameter of right subtree
-	path = lh+rh+1  #Longest path between leaves that goes through the node (tree)
-	d = max([ld, rd, path])
+	longest_path = lh+rh+1                #Longest path between leaves that goes through the node (tree)
 	h = max(lh,rh) + 1
-	#print("Node(%d) diameter=%d height=%d" % (tree.data,d,h))
+	d = max([ld, rd, longest_path])
+	#print("Node(%d) height=%d diameter=%d" % (tree.data,h,d))
 	return (d,h)	
 
-
+'''
+Example 1: diameter via root
+                         1
+                       /   \ 
+                      2     3
+                    /   \
+                  4      5
+ 
+ node[4]: height=1, diameter=1
+ node[5]: height=1, diameter=1
+ node[2]: height=2, diamter= max(1,1,1+1+1) = 3
+ node[3]: height=1, diameter=1
+ node[1]: height=max(2,1)+1 = 3, diameter=max(3,1,2+1+1) = 4
+'''
+#Example 1
 root = Tree.BinaryTree(1)
 root.insertLeft(2)
 root.insertRight(3)
 root.getLeft().insertLeft(4)
 root.getLeft().insertRight(5)
-#root.getRight().insertLeft(6)
-#root.getRight().insertRight(7)
-#root.getLeft().getLeft().insertLeft(8)
-#root.getLeft().getRight().insertRight(9)
-#root.getLeft().getRight().getRight().insertRight(10)
-#root.getLeft().getRight().getRight().getRight().insertRight(11)
+d,h = diameter_height(root)
+print("Diameter of tree 1 (time complexity O(n)) = %d and max height=%d" % (d,h))
 
-print("Diameter of tree (time complexity O(n^2)) = %d" % diameter(root))
+'''
+Example 2: diameter not via root
+						  1
+						/   
+					   2	
+					 /   \
+				    4     5
+				  /        \
+				 6          7
+				             \
+				              8   
+node[8]: height=1 , diameter=1 
+node[7]: height=2 , diameter=2 
+node[5]: height=3 , diameter=3 
+node[6]: height=1 , diameter=1 
+node[4]: height=1 , diameter=2 
+node[2]: height=4 , diameter=max(2,3,2+3+1)=6 
+node[1]: height=5 , diameter=max(6,0,4+0+1)=6 
+'''
+root = None
+root = Tree.BinaryTree(1)
+root.insertLeft(2)
+root.getLeft().insertLeft(4)
+root.getLeft().insertRight(5)
+root.getLeft().getLeft().insertLeft(6)
+root.getLeft().getRight().insertRight(7)
+root.getLeft().getRight().getRight().insertRight(8)
 d,h = diameter_height(root)
-print("Diameter of tree (time complexity O(n)) = %d and max height=%d" % (d,h))
+print("Diameter of tree 2 (time complexity O(n)) = %d and max height=%d" % (d,h))