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Merge pull request #2759 from miaoxu404/master

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Update 0513.找树左下角的值.md
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程序员Carl
2024-10-16 10:18:20 +08:00
committed by GitHub
parent 42d84f8a7f d0f6653a6f
commit 371cfe182a
2 changed files with 11 additions and 11 deletions
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20
problems/0112.路径总和.md
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@ -564,10 +564,10 @@ class Solution:
return False return False
def hasPathSum(self, root: TreeNode, sum: int) -> bool: def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
if root is None: if root is None:
return False return False
return self.traversal(root, sum - root.val) return self.traversal(root, targetSum - root.val)
``` ```
(版本二) 递归 + 精简 (版本二) 递归 + 精简
@ -579,12 +579,12 @@ class Solution:
# self.left = left # self.left = left
# self.right = right # self.right = right
class Solution: class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool: def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
if not root: if not root:
return False return False
if not root.left and not root.right and sum == root.val: if not root.left and not root.right and targetSum == root.val:
return True return True
return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val) return self.hasPathSum(root.left, targetSum - root.val) or self.hasPathSum(root.right, targetSum - root.val)
``` ```
(版本三) 迭代 (版本三) 迭代
@ -596,7 +596,7 @@ class Solution:
# self.left = left # self.left = left
# self.right = right # self.right = right
class Solution: class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool: def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
if not root: if not root:
return False return False
# 此时栈里要放的是pair<节点指针，路径数值> # 此时栈里要放的是pair<节点指针，路径数值>
@ -659,13 +659,13 @@ class Solution:
return return
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]: def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
self.result.clear() self.result.clear()
self.path.clear() self.path.clear()
if not root: if not root:
return self.result return self.result
self.path.append(root.val) # 把根节点放进路径 self.path.append(root.val) # 把根节点放进路径
self.traversal(root, sum - root.val) self.traversal(root, targetSum - root.val)
return self.result return self.result
``` ```
@ -678,7 +678,7 @@ class Solution:
# self.left = left # self.left = left
# self.right = right # self.right = right
class Solution: class Solution:
def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]: def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
result = [] result = []
self.traversal(root, targetSum, [], result) self.traversal(root, targetSum, [], result)
@ -703,7 +703,7 @@ class Solution:
# self.left = left # self.left = left
# self.right = right # self.right = right
class Solution: class Solution:
def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]: def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
if not root: if not root:
return [] return []
stack = [(root, [root.val])] stack = [(root, [root.val])]

2
problems/0513.找树左下角的值.md
View File

@ -55,7 +55,7 @@
参数必须有要遍历的树的根节点，还有就是一个int型的变量用来记录最长深度。 这里就不需要返回值了，所以递归函数的返回类型为void。 参数必须有要遍历的树的根节点，还有就是一个int型的变量用来记录最长深度。 这里就不需要返回值了，所以递归函数的返回类型为void。
本题还需要类里的两个全局变量，maxLen用来记录最大深度，result记录最大深度最左节点的数值。 本题还需要类里的两个全局变量，maxDepth用来记录最大深度，result记录最大深度最左节点的数值。
代码如下： 代码如下：