LeetCode #617 合并二叉树 树 递归
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2022-03-03 10:35:41
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LeetCode #617 合并二叉树
题目描述
给定两个二叉树,想象当你将它们中的一个覆盖到另一个上时,两个二叉树的一些节点便会重叠。
你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠,那么将他们的值相加作为节点合并后的新值,否则 不为 NULL 的节点将直接作为新二叉树的节点。
示例 1:
输入:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
输出:
合并后的树:
3
/ \
4 5
/ \ \
5 4 7
注意: 合并必须从两个树的根节点开始。
方法一:递归
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
if not t1 and t2:
return t2
elif t1 and t2:
t1.val = t1.val + t2.val
t1.left = self.mergeTrees(t1.left, t2.left)
t1.right = self.mergeTrees(t1.right, t2.right)
# t1 and not t2
# else:
# return t1
return t1
- 时间复杂度:
- 空间复杂度:
方法二:递归
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
if not t1 and t2:
return t2
elif t1 and t2:
stack = [(t1, t2)]
while stack:
r1, r2 = stack.pop(0)
r1.val += r2.val
# 如果r1和r2的左子树都存在就把r2的左子树挂在r1上
if r1.left and r2.left:
stack.append((r1.left, r2.left))
elif not r1.left:
r1.left = r2.left
# 右子树也是一样的
if r1.right and r2.right:
stack.append((r1.right, r2.right))
elif not r1.right:
r1.right = r2.right
# t1 and not t2
# else:
# return t1
return t1
- 时间复杂度:
- 空间复杂度:
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