kth node with of BT during inorder traversal

Author: david-legend

src/EPI_prep/binary_trees/0_binary_tree_traversal/tree_traversals.py renamed to EPI_prep/src/binary_trees/0_binary_tree_traversal/tree_traversals.py

@@ -0,0 +1,81 @@
+from collections import namedtuple
+
+
+class Node:
+    def __init__(self, val):
+        self.val = val
+        self.left = self.right = None
+
+
+def kth_node(tree, k):
+    Result = namedtuple('Result', ['status', 'node'])
+    i = 0
+
+    def inorder(node):
+        nonlocal i
+        if node is None:
+            return Result(False, None)
+
+        left_result = inorder(node.left)
+        if left_result.status:
+            return left_result
+
+        i += 1
+        if i == k:
+            return Result(True, node)
+        
+        right_result = inorder(node.right)
+        if right_result.status:
+            return right_result
+        
+        return Result(False, None)
+
+    
+    return inorder(tree).node
+
+def kth_node_2(tree, k):
+    result = []
+
+    def inorder(node):
+        if node is None:
+            return
+        inorder(node.left)
+        result.append(node)
+        inorder(node.right)
+
+    inorder(tree)
+    return result[k-1] if k <= len(result) else None
+
+
+    
+
+node_1 = Node(1)
+node_2 = Node(2)
+node_3 = Node(3)
+node_4 = Node(4)
+node_5 = Node(5)
+node_6 = Node(6)
+node_7 = Node(7)
+
+node_1.left, node_1.right = node_2, node_3
+node_2.left, node_2.right = node_4, node_5
+node_3.left, node_3.right = node_6, node_7
+
+
+print("\nKth Node First Solution\n")
+print(kth_node(node_1, 1)) #4
+print(kth_node(node_1, 2)) #2
+print(kth_node(node_1, 3)) #5
+print(kth_node(node_1, 4)) #1
+print(kth_node(node_1, 5)) #6
+print(kth_node(node_1, 6)) #3
+print(kth_node(node_1, 7)) #7
+
+print("\nKth Node Non Optimal Solution\n")
+print(kth_node_2(node_1, 1)) #4
+print(kth_node_2(node_1, 2)) #2
+print(kth_node_2(node_1, 3)) #5
+print(kth_node_2(node_1, 4)) #1
+print(kth_node_2(node_1, 5)) #6
+print(kth_node_2(node_1, 6)) #3
+print(kth_node_2(node_1, 7)) #7

src/EPI_prep/binary_trees/1_is_binary_tree_height_balanced/README.md renamed to EPI_prep/src/binary_trees/1_is_binary_tree_height_balanced/README.md

@@ -0,0 +1,41 @@
+class Node:
+    def __init__(self, val, size = 0):
+        self.val, self.size = val, size
+        self.left = self.right = None
+
+
+def kth_node(tree, k):
+    if tree:
+        while tree:
+            left_size = tree.left.size if tree.left else 0
+            if left_size + 1 < k:
+                tree = tree.right
+                k -= left_size + 1
+            elif left_size == k - 1:
+                return tree.val
+            else:
+                tree = tree.left
+    
+    return None
+    
+
+node_1 = Node(1, 7)
+node_2 = Node(2, 3)
+node_3 = Node(3, 3)
+node_4 = Node(4, 1)
+node_5 = Node(5, 1)
+node_6 = Node(6, 1)
+node_7 = Node(7, 1)
+
+node_1.left, node_1.right = node_2, node_3
+node_2.left, node_2.right = node_4, node_5
+node_3.left, node_3.right = node_6, node_7
+
+
+print(kth_node(node_1, 1))
+print(kth_node(node_1, 2))
+print(kth_node(node_1, 3))
+print(kth_node(node_1, 4))
+print(kth_node(node_1, 5))
+print(kth_node(node_1, 6))
+print(kth_node(node_1, 7))