algorithm/leetcode-master
1
0
Fork 0
mirror of https://github.com/youngyangyang04/leetcode-master.git synced 2025-07-15 07:20:40 +08:00
Code Issues Packages Projects Releases Wiki Activity

Update 0404.左叶子之和.md

Browse Source 
勘误python注释
This commit is contained in:
Asterisk
2021-10-08 13:12:08 +08:00
committed by GitHub
parent aea3faf3b7
commit 5e5b3d5f4c
1 changed files with 4 additions and 31 deletions
Download Patch File Download Diff File Expand all files Collapse all files

35
problems/0404.左叶子之和.md
View File

@ -171,10 +171,10 @@ class Solution {
int rightValue = sumOfLeftLeaves(root.right); // 右
int midValue = 0;
if (root.left != null && root.left.left == null && root.left.right == null) { // 中
if (root.left != null && root.left.left == null && root.left.right == null) {
midValue = root.left.val;
}
int sum = midValue + leftValue + rightValue;
int sum = midValue + leftValue + rightValue; // 中
return sum;
}
}
@ -242,37 +242,10 @@ class Solution:
cur_left_leaf_val = 0
if root.left and not root.left.left and not root.left.right:
cur_left_leaf_val = root.left.val # 中
cur_left_leaf_val = root.left.val
return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum
return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum # 中
```
> 递归前序遍历
```python3
# 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 sumOfLeftLeaves(self, root: TreeNode) -> int:
# 需要通过中节点来判断其的左节点是否存在；左节点自己的左右节点也是否存在
if not root: return 0
# 初始化left_leaf备用
left_leaf = 0
# 若当前节点的左孩子就是左叶子
if root.left and not root.left.left and not root.left.right:
left_leaf = root.left.val
left_left_leaves_sum = self.sumOfLeftLeaves(root.left)
right_left_leaves_sum = self.sumOfLeftLeaves(root.right)
return left_leaf + left_left_leaves_sum + right_left_leaves_sum
```
> 迭代
```python3