题目:原题链接(中等)
标签:树、二叉树、深度优先搜索
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
---|---|---|---|
Ans 1 (Python) | 44ms (60.76%) | ||
Ans 2 (Python) | 40ms (82.28%) | ||
Ans 3 (Python) |
解法一(递归):
class Solution:
def sumNumbers(self, root: TreeNode, now_val=0) -> int:
# 处理当前节点不存在的情况
if not root:
return 0
new_val = now_val * 10 + root.val
# 处理当前节点非叶节点的情况
if root.left or root.right:
return self.sumNumbers(root.left, new_val) + self.sumNumbers(root.right, new_val)
# 处理当前节点为叶节点的情况
else:
return new_val
解法二(优化解法一):
class Solution:
def __init__(self):
self.ans = 0
def sumNumbers(self, root: TreeNode) -> int:
self.count(root)
return self.ans
def count(self, node, now_val=0):
if node:
new_val = now_val * 10 + node.val
if not node.left and not node.right:
self.ans += new_val
self.count(node.left, new_val)
self.count(node.right, new_val)