原始题目

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

题目分析

判断一个二元树的左子叶总和

把每个节点送进递回函数里面取得该节点的左节点，并且送进同一个函数取得更深入的左节点
如果传入节点的左节点不为空，且该节点的左节点没有更深入的左节点
而且该节点的左节点的右节点为空，则找到了一个左子叶

判断式长得像：

if (node->left != NULL && node->left->left == NULL && node->left->right == NULL) {
    找到左子叶
}

解题过程

# 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

结果