Leetcode 129. Sum Root to Leaf Numbers

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于 2021-11-28 18:23:12 发布

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分类专栏： Leetcode 解题报告文章标签： leetcode 算法

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这篇博客讲解了如何通过递归算法计算给定二叉树中每个从根到叶节点路径表示的数字之和，解决了一个涉及树遍历和数值计算的问题。算法适用于树深度不超过10且节点值在0-9之间的场景，适用于32位整数范围内的答案。

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Problem

You are given the root of a binary tree containing digits from 0 to 9 only.
Each root-to-leaf path in the tree represents a number.

For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123.

Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.
A leaf node is a node with no children.

Constraints:

The number of nodes in the tree is in the range [1, 1000].
0 <= Node.val <= 9
The depth of the tree will not exceed 10.

Algorithm

Recursion.

Code

# 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 sumNumbers(self, root: Optional[TreeNode]) -> int:
        ans = 0
        def dfs(root, now):
            nonlocal ans
            if not root:
                return
            elif not root.right and not root.left:
                ans += now * 10 + root.val
            elif not root.right:
                dfs(root.left, now*10 + root.val)
            elif not root.left:
                dfs(root.right, now*10 + root.val)
            else:
                dfs(root.left, now*10 + root.val)
                dfs(root.right, now*10 + root.val)
                
        dfs(root, 0)
        
        return ans