617.py

# leetcode 617 合并二叉树


# 首先是递归法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:

        ## 首先我们尝试使用递归去做
        if not root1: return root2
        if not root2: return root1

        root1.val += root2.val

        root1.left = self.mergeTrees(root1.left, root2.left)
        root1.right = self.mergeTrees(root1.right, root2.right)

        return root1



# 然后是迭代法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:

        # 迭代法
        if not root2 and not root1:
            return None
        if not root1: return root2
        if not root2: return root1

        queue = collections.deque([root1, root2])

        while queue:
            node1, node2 = queue.popleft(), queue.popleft()

            if node1.left and node2.left:
                queue.append(node1.left)
                queue.append(node2.left)

            if node1.right and node2.right:
                queue.append(node1.right)
                queue.append(node2.right)

            node1.val += node2.val

            if not node1.left and node2.left:
                node1.left = node2.left

            if not node1.right and node2.right:
                node1.right = node2.right

        return root1