404. 左叶子之和

Difficulty

easy

分析

使用简单的二叉树遍历，前序中序后序均可以，在访问某节点时进行左叶子节点判断 node.left is None and node.left.left is None and node.lrft.right is None 如果是左叶子节点，则将该左叶子节点值加到结果中

法1 递归法

将结果定义为全局变量，递归程序这样只负责遍历，遇到左叶子节点就将


# 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 __init__(self):
        self.sum = 0
    def sumOfLeftLeaves(self, root: TreeNode) -> int:
        self.recur(node=root)
        return self.sum

    def recur(self, node):
        if node is None:
            return
        if node.left and node.left.left is None and node.left.right is None:
            self.sum += node.left.val
        self.sumOfLeftLeaves(node.left)
        self.sumOfLeftLeaves(node.right)

法2 信息传递

将结果参与到递归遍历中，有种信息传递的感觉


# 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:
        if root is None:
            return 0
        sum = 0
        if root.left and root.left.left is None and root.left.right is None:
            sum += root.left.val
        leftSum = self.sumOfLeftLeaves(root.left)
        rightSum = self.sumOfLeftLeaves(root.right)
        return sum + leftSum + rightSum