LeetCode 617. Merge Two Binary Trees - Python

LeetCode 617. Merge Two Binary Trees - Python

 

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

 

Example 1:

Input: 
	Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  

Output: 
Merged tree:
	     3
	    / \
	   4   5
	  / \   \ 
	 5   4   7

Note: The merging process must start from the root nodes of both trees.

 

 

Solution 1 (Recusive) :

class Solution:
    def mergeTrees(self, t1, t2):
        """
        :type t1: TreeNode
        :type t2: TreeNode
        :rtype: TreeNode
        """
        if t1 is None:
            return t2
        if t2 is None:
            return t1
        t1.val = t1.val + t2.val
        t1.left = self.mergeTrees(t1.left, t2.left)
        t1.right = self.mergeTrees(t1.right, t2.right)
        return t1

 

 

Solution 2 (Recusive) :

class Solution:
    def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
        if not t1 and not t2:
            return None
        if bool(t1) ^ bool(t2):
            return t1 if t1 else t2
        t1.val += t2.val
        t1.left = self.mergeTrees(t1.left, t2.left)
        t1.right = self.mergeTrees(t1.right, t2.right)
        return t1

 

Solution 3 (Iterative) :

class Solution:
    def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
        if not t1 and not t2:
            return None
        if bool(t1) ^ bool(t2):
            return t1 if t1 else t2
        queue = deque([(t1, t2)])
        while queue:
            tree1, tree2 = queue.popleft()
            tree1.val += tree2.val
            if not tree1.left and tree2.left:
                tree1.left = tree2.left
            elif tree1.left and tree2.left:
                queue.append((tree1.left, tree2.left))
            if not tree1.right and tree2.right:
                tree1.right = tree2.right
            elif tree1.right and tree2.right:
                queue.append((tree1.right, tree2.right))
        return t1

 

 

 

 

관련 개념 :

재귀함수

재귀와 반복

비트연산자 - xor 연산자

docstring

magic method : __init__,  __add__, __doc__

동적계획법

queue. pop(0) 함수

DFS, BFS

큐(queue) 자료구조