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更改 0112.路径总和.md Python3 代码.

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0112.路径总和:
1. 修改了 python3 注释, 更改代码格式.
2. 增加了迭代方法Python3代码.

0113.路径总和-ii:
上一版python3代码的问题: (1)注释一般不用// (2)代码格式有些不规范. (3)内层函数命名也不是特别的informative.

新版本代码修改了上述问题.

本次提交代码均已通过leetcode测试.
This commit is contained in:
Kelvin
2021-07-29 15:50:46 -04:00
parent ff3875d502
commit 440e04fffe
1 changed files with 72 additions and 41 deletions
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113
problems/0112.路径总和.md
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@ -416,6 +416,8 @@ class Solution {
Python：
0112.路径总和
**递归**
```python
# Definition for a binary tree node.
# class TreeNode:
@ -424,28 +426,56 @@ Python：
# self.left = left
# self.right = right
// 递归法
class Solution:
def hasPathSum(self, root: TreeNode, targetSum: int) -> bool:
def isornot(root,targetSum)->bool:
if (not root.left) and (not root.right) and targetSum == 0:return True // 遇到叶子节点，并且计数为0
if (not root.left) and (not root.right):return False //遇到叶子节点，计数不为0
def isornot(root, targetSum) -> bool:
if (not root.left) and (not root.right) and targetSum == 0:
return True # 遇到叶子节点，并且计数为0
if (not root.left) and (not root.right):
return False # 遇到叶子节点，计数不为0
if root.left:
targetSum -= root.left.val //左节点
if isornot(root.left,targetSum):return True //递归，处理左节点
targetSum += root.left.val //回溯
targetSum -= root.left.val # 左节点
if isornot(root.left, targetSum): return True # 递归，处理左节点
targetSum += root.left.val # 回溯
if root.right:
targetSum -= root.right.val //右节点
if isornot(root.right,targetSum):return True //递归，处理右节点
targetSum += root.right.val //回溯
targetSum -= root.right.val # 右节点
if isornot(root.right, targetSum): return True # 递归，处理右节点
targetSum += root.right.val # 回溯
return False
if root == None:return False //别忘记处理空TreeNode
else:return isornot(root,targetSum-root.val)
if root == None:
return False # 别忘记处理空TreeNode
else:
return isornot(root, targetSum - root.val)
```
**迭代 - 层序遍历**
```python
class Solution:
def hasPathSum(self, root: TreeNode, targetSum: int) -> bool:
if not root:
return False
stack = [] # [(当前节点，路径数值), ...]
stack.append((root, root.val))
while stack:
cur_node, path_sum = stack.pop()
if not cur_node.left and not cur_node.right and path_sum == targetSum:
return True
if cur_node.right:
stack.append((cur_node.right, path_sum + cur_node.right.val))
if cur_node.left:
stack.append((cur_node.left, path_sum + cur_node.left.val))
return False
```
0113.路径总和-ii
**递归**
```python
# Definition for a binary tree node.
# class TreeNode:
@ -453,35 +483,36 @@ class Solution:
# self.val = val
# self.left = left
# self.right = right
//递归法
class Solution:
def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
path=[]
res=[]
def pathes(root,targetSum):
if (not root.left) and (not root.right) and targetSum == 0: // 遇到叶子节点，并且计数为0
res.append(path[:]) //找到一种路径，记录到res中，注意必须是path[:]而不是path
return
if (not root.left) and (not root.right):return // 遇到叶子节点直接返回
if root.left: //左
targetSum -= root.left.val
path.append(root.left.val) //递归前记录节点
pathes(root.left,targetSum) //递归
targetSum += root.left.val //回溯
path.pop() //回溯
if root.right: //右
targetSum -= root.right.val
path.append(root.right.val) //递归前记录节点
pathes(root.right,targetSum) //递归
targetSum += root.right.val //回溯
path.pop() //回溯
return
if root == None:return [] //处理空TreeNode
else:
path.append(root.val) //首先处理根节点
pathes(root,targetSum-root.val)
return res
def traversal(cur_node, remain):
if not cur_node.left and not cur_node.right and remain == 0:
result.append(path[:])
return
if not cur_node.left and not cur_node.right: return
if cur_node.left:
path.append(cur_node.left.val)
remain -= cur_node.left.val
traversal(cur_node.left, remain)
path.pop()
remain += cur_node.left.val
if cur_node.right:
path.append(cur_node.right.val)
remain -= cur_node.right.val
traversal(cur_node.right, remain)
path.pop()
remain += cur_node.right.val
result, path = [], []
if not root:
return []
path.append(root.val)
traversal(root, targetSum - root.val)
return result
```
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