build tree & sum path (usage of list[0]) - 404. Sum of Left Leaves

本文介绍了一种使用广度优先搜索构建二叉树的方法,并实现了一个核心算法来计算所有左叶子节点的总和。通过具体实例展示了如何从列表中构建二叉树,然后遍历该树并计算所有左叶子节点的值之和。

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from collections import deque
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def printTreeCore(self, root, preorder=True):
        if not root:
            return
        print('{}->'.format(root.val), end='')
        self.printTreeCore(root.left)
        self.printTreeCore(root.right)
        return

    def printTree(self, root, preorder=True):
        self.printTreeCore(root)
        print("\b\b")
        return

    def buildtree_breadth(self, breadthlist):
        if not breadthlist:
            return None

        root = TreeNode(breadthlist.pop(0))
        q = deque([root])
        while breadthlist:
            node = q.popleft()
            val = breadthlist.pop(0)
            if val is not None:
                node.left = TreeNode(val)
            if not breadthlist:
                break
            val = breadthlist.pop(0)
            if val is not None:
                node.right = TreeNode(val)
            if node.left:
                q.append(node.left)
            if node.right:
                q.append(node.right)
        return root
        
    def get_left_leave(self, node, flag, l):
        if node is None:
            return
        if flag == 1 and node.left is None and node.right is None:
            l[0] += node.val
        self.get_left_leave(node.left, 1, l)
        self.get_left_leave(node.right, 0, l)
        return

    def sumOfLeftLeaves_leetcode(self, root: TreeNode) -> int:
        sum = [0]
        self.get_left_leave(root, 0, sum)
        return sum[0]


if __name__ == "__main__":
    s = Solution()
    root = s.buildtree_breadth([3, 9, 20, None, None, 15, 7])
    s.printTree(root)
    res = s.sumOfLeftLeaves(root)
    print(res)
    
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