题解 | #合并二叉树#

递归法 - 前序遍历

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param t1 TreeNode类 
# @param t2 TreeNode类 
# @return TreeNode类
#
class Solution:
    def mergeTrees(self , root1: TreeNode, root2: TreeNode) -> TreeNode:
        # 递归终止条件: 
        #  但凡有一个节点为空, 就立刻返回另外一个. 如果另外一个也为None就直接返回None. 
        if not root1: 
            return root2
        if not root2: 
            return root1
        # 上面的递归终止条件保证了代码执行到这里root1, root2都非空. 
        root1.val += root2.val # 中
        root1.left = self.mergeTrees(root1.left, root2.left) #左
        root1.right = self.mergeTrees(root1.right, root2.right) # 右
        
        return root1 # ⚠️ 注意: 本题我们重复使用了题目给出的节点而不是创建新节点. 节省时间, 空间. 

迭代法

class Solution:
    def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
        if not root1: 
            return root2
        if not root2: 
            return root1

        queue = deque()
        queue.append(root1)
        queue.append(root2)

        while queue: 
            node1 = queue.popleft()
            node2 = queue.popleft()
            # 更新queue
            # 只有两个节点都有左节点时, 再往queue里面放.
            if node1.left and node2.left: 
                queue.append(node1.left)
                queue.append(node2.left)
            # 只有两个节点都有右节点时, 再往queue里面放.
            if node1.right and node2.right: 
                queue.append(node1.right)
                queue.append(node2.right)

            # 更新当前节点. 同时改变当前节点的左右孩子. 
            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