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94 lines
2.2 KiB
Python
94 lines
2.2 KiB
Python
#!/usr/local/bin/python3
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"""
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Problem Description: Given two binary tree, return the merged tree.
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The rule for merging is that if two nodes overlap, then put the value sum of
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both nodes to the new value of the merged node. Otherwise, the NOT null node
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will be used as the node of new tree.
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"""
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from __future__ import annotations
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class Node:
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"""
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A binary node has value variable and pointers to its left and right node.
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"""
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def __init__(self, value: int = 0) -> None:
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self.value = value
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self.left: Node | None = None
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self.right: Node | None = None
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def merge_two_binary_trees(tree1: Node | None, tree2: Node | None) -> Node | None:
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"""
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Returns root node of the merged tree.
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>>> tree1 = Node(5)
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>>> tree1.left = Node(6)
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>>> tree1.right = Node(7)
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>>> tree1.left.left = Node(2)
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>>> tree2 = Node(4)
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>>> tree2.left = Node(5)
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>>> tree2.right = Node(8)
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>>> tree2.left.right = Node(1)
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>>> tree2.right.right = Node(4)
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>>> merged_tree = merge_two_binary_trees(tree1, tree2)
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>>> print_preorder(merged_tree)
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9
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11
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2
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1
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15
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4
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"""
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if tree1 is None:
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return tree2
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if tree2 is None:
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return tree1
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tree1.value = tree1.value + tree2.value
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tree1.left = merge_two_binary_trees(tree1.left, tree2.left)
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tree1.right = merge_two_binary_trees(tree1.right, tree2.right)
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return tree1
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def print_preorder(root: Node | None) -> None:
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"""
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Print pre-order traversal of the tree.
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>>> root = Node(1)
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>>> root.left = Node(2)
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>>> root.right = Node(3)
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>>> print_preorder(root)
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1
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2
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3
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>>> print_preorder(root.right)
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3
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"""
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if root:
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print(root.value)
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print_preorder(root.left)
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print_preorder(root.right)
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if __name__ == "__main__":
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tree1 = Node(1)
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tree1.left = Node(2)
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tree1.right = Node(3)
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tree1.left.left = Node(4)
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tree2 = Node(2)
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tree2.left = Node(4)
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tree2.right = Node(6)
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tree2.left.right = Node(9)
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tree2.right.right = Node(5)
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print("Tree1 is: ")
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print_preorder(tree1)
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print("Tree2 is: ")
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print_preorder(tree2)
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merged_tree = merge_two_binary_trees(tree1, tree2)
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print("Merged Tree is: ")
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print_preorder(merged_tree)
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