心路历程:
看到两颗二叉树的问题,第一反应想到了同频遍历,然后每一步创建新的结点,虽然也写出来了但是代码比较长,而且空间复杂度比较高,好处是没有修改原始的两个二叉树的结果。
后来看了网上的解答,发现可以直接按照修改其中一个二叉树来做,这样代码量会少很多,并且空间复杂度会低,但是这样会导致原来的二叉树被修改。
注意的点:
1、每当调用node.left时都需要保证node不是None
2、当想不明白应该在循环哪一步新建变量时,可以按照循环不变量的角度去思考
3、递归函数return值时,边界return,非边界也要return,是统一的
解法一:同频遍历+创建新的二叉树
# 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 and not root2:
return None
new_root = TreeNode()
# 二叉树同频深度遍历
def dfs(node1, node2, new_node):
if not node1 and not node2: return
if node1 and not node2:
new_node.val = node1.val
elif not node1 and node2:
new_node.val = node2.val
else:
new_node.val = node1.val + node2.val
if (node1 and node1.left) or (node2 and node2.left):
new_node_left = TreeNode()
new_node.left = new_node_left
if node1 and node2:
dfs(node1.left, node2.left, new_node.left)
elif node1 and not node2:
dfs(node1.left, None, new_node.left)
elif not node1 and node2:
dfs(None, node2.left, new_node.left)
if (node1 and node1.right) or (node2 and node2.right):
new_node_right = TreeNode()
new_node.right = new_node_right
if node1 and node2:
dfs(node1.right, node2.right, new_node.right)
elif node1 and not node2:
dfs(node1.right, None, new_node.right)
elif not node1 and node2:
dfs(None, node2.right, new_node.right)
dfs(root1, root2, new_root)
return new_root
解法二:修改其中一颗二叉树,递归函数直接返回合并后的一个结点
def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:
def dfs(node1, node2): # 将node2结点合并到node1,返回合并后的node1结点
if not node1: return node2
if not node2: return node1
node1.val += node2.val
node1.left = dfs(node1.left, node2.left)
node1.right = dfs(node1.right, node2.right)
return node1
return dfs(root1, root2)