[LeetCode] #404 Sum of Left Leaves (Easy)
LeetCode 第 404 題 Sum of Left Leaves，難度 Easy
用 Python3 解 LeetCode 系列，Sum of Left Leaves，屬於 Easy
原始題目
Given the root of a binary tree, return the sum of all left leaves.
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: 24
Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.
Example 2:
Input: root = [1]
Output: 0
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. -1000 <= Node.val <= 1000
題目分析
判斷一個二元樹的左子葉總和
把每個節點送進遞迴函數裡面取得該節點的左節點，並且送進同一個函數取得更深入的左節點
如果傳入節點的左節點不為空，且該節點的左節點沒有更深入的左節點
而且該節點的左節點的右節點為空，則找到了一個左子葉
判斷式長得像：
if (node->left != NULL && node->left->left == NULL && node->left->right == NULL) {
找到左子葉
}
解題過程
# 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: Optional[TreeNode]) -> int:
if not root:
return 0
left_left_leaves_sum = self.sumOfLeftLeaves(root.left)
right_left_leaves_sum = self.sumOfLeftLeaves(root.right)
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
return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum
結果
