update children count and remove which child

laurentluce · laurentluce · commit 3c41ce14ca58 · 2010-12-21T11:54:25.000-08:00
diff --git a/algorithms/binary_tree.py b/algorithms/binary_tree.py
@@ -18,7 +18,7 @@ def insert(self, data):
 
     @param data node data object to insert
     """
-    elif data < self.data:
+    if data < self.data:
       if self.left == None:
         self.left = Node(data)
       else:
@@ -57,10 +57,10 @@ def delete(self, data):
     # get node containing data
     node, parent = self.lookup(data)
     if node != None:
-      children_count = self.children_count(node)
+      children_count = node.children_count()
       if children_count == 0:
         # if node has no children, just remove it
-        if self.which_child(parent, node) == 'left':
+        if parent.left is node:
           parent.left = None
         else:
           parent.right = None
@@ -139,33 +139,16 @@ def tree_data(self):
         yield node.data
         node = node.right
 
-  def which_child(self, parent, child):
-    """
-    Test if child is parent's left child or parent's right child
-
-    @param parent parent node
-    @param child parent's child node
-    """
-    if parent == None:
-      return None
-    if parent.left == child:
-      return 'left'
-    else:
-      return 'right'
-
-  def children_count(self, node):
+  def children_count(self):
     """
     Return the number of children
 
-    @param node node to get nb of children
     @returns number of children: 0, 1, 2
     """
-    if node == None:
-      return None
     cnt = 0
-    if node.left:
+    if self.left:
       cnt += 1
-    if node.right:
+    if self.right:
       cnt += 1
     return cnt
 
diff --git a/binary_tree_tutorial.txt b/binary_tree_tutorial.txt
@@ -197,7 +197,7 @@ class Node:
     # get node containing data
     node, parent = self.lookup(data)
     if node != None:
-      children_count = self.children_count(node)
+      children_count = node.children_count()
     ...
 [/code]
 
@@ -215,57 +215,36 @@ Let's tackle the first possibility which is the easiest. We look for the node to
     ...
     if children_count == 0:
       # if node has no children, just remove it
-      if self.which_child(parent, node) == 'left':
+      if parent.left is node:
         parent.left = None
       else:
         parent.right = None
     ...
 [/code]
 
-Note: children_count() returns the number of children of a node. which_child() tells if the node passed is the left child or right child of parent.
+Note: children_count() returns the number of children of a node.
 
 Here is the function children_count:
 
 [code lang="python"]
 class Node:
   ...
-  def children_count(node):
+  def children_count(self):
     """
     Returns the number of children
 
-    @param node node to get nb of children
     @returns number of children: 0, 1, 2
     """
     if node == None:
       return None
     cnt = 0
-    if node.left:
+    if self.left:
       cnt += 1
-    if node.right:
+    if self.right:
       cnt += 1
     return cnt
 [/code]
 
-And which_child():
-
-[code lang="python"]
-class Node:
-  ...
-  def which_child(parent, child):
-    """
-    Test if child is parent's left child or parent's right child
-
-    @param parent parent node
-    @param child parent's child node
-    """
-    if parent == None:
-      return None
-    if parent.left == child:
-      return 'left'
-    else:
-      return 'right'
-[/code]
-
 For example, we want to remove node 1. Node 3 left child will be set to None.
 
 [code lang="python"]
