404. 左叶子之和

题目

404. 左叶子之和

找到左叶子

方法一：递归

1. 判断根节点是否有左子树，用sum_left记录
2. 递归左子树与右子树，找出其中的叶子节点

# 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:
        sum_left = 0
        if not root:
            return 0
        if root.left and not (root.left.left or root.left.right):# 是否为左叶子节点
                sum_left = root.left.val
        return sum_left + self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)

方法二 栈

1. 求出所有叶子点的和
2. 减去右叶子节点的和
   在通过栈遍历的基础上修改的

# 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 not root:
            return 0
        sum_left = 0
        stack = []
        last_visit = root
       
        while root or stack:
            while root:
                # print(root.val)# 先序
                stack.append(root)
                root = root.left
            root = stack[-1]

            if root.right != last_visit:
                # print(root.val)# 中序
                root = root.right
                if root and not (root.left or root.right):# 所有右叶子节点的和
                    sum_left -= root.val
                if not root:
                    last_visit = None
            else:
                temp = stack.pop()
                if not (temp.left or temp.right):# 这里加的是所有叶子节点的和
                    sum_left += root.val
                # print(temp.val)# 后序
                last_visit = temp
                root = None
        return sum_left