LeetCode617. 合并二叉树
- 题目链接
- 代码
题目链接
https://leetcode.cn/problems/merge-two-binary-trees/
代码
递归
# 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]:
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
迭代
# 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]:
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()
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