129 Sum Root to Leaf Numbers 求根到叶子节点数字之和
Description:
Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
Note:
A leaf is a node with no children.
Example:
Example 1:
Input: [1,2,3]
1
/ \
2 3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:
Input: [4,9,0,5,1]
4
/ \
9 0
/ \
5 1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
题目描述:
给定一个二叉树，它的每个结点都存放一个 0-9 的数字，每条从根到叶子节点的路径都代表一个数字。
例如，从根到叶子节点路径 1->2->3 代表数字 123。
计算从根到叶子节点生成的所有数字之和。
说明：
叶子节点是指没有子节点的节点。
示例 :
示例 1:
输入: [1,2,3]
1
/ \
2 3
输出: 25
解释:
从根到叶子节点路径 1->2 代表数字 12.
从根到叶子节点路径 1->3 代表数字 13.
因此，数字总和 = 12 + 13 = 25.
示例 2:
输入: [4,9,0,5,1]
4
/ \
9 0
/ \
5 1
输出: 1026
解释:
从根到叶子节点路径 4->9->5 代表数字 495.
从根到叶子节点路径 4->9->1 代表数字 491.
从根到叶子节点路径 4->0 代表数字 40.
因此，数字总和 = 495 + 491 + 40 = 1026.
思路:
参考LeetCode #112 Path Sum 路径总和
不过这里的路径需要每次递归到新的一层时✖️10
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
public:
int sumNumbers(TreeNode* root)
{
return helper(root, 0);
}
private:
int helper(TreeNode* root, int path)
{
if (!root) return 0;
int result = path * 10 + root -> val;
if (!root -> left and !root -> right) return result;
return helper(root -> left, result) + helper(root -> right, result);
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumNumbers(TreeNode root) {
return helper(root, 0);
}
private int helper(TreeNode root, int path) {
if (root == null) return 0;
int result = path * 10 + root.val;
if (root.left == null && root.right == null) return result;
return helper(root.left, result) + helper(root.right, result);
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def sumNumbers(self, root: TreeNode) -> int:
f = lambda root: [] if not root else [root.val] if not root.left and not root.right else [str(root.val) + str(i) for i in f(root.left) + f(root.right)]
return sum(map(int, f(root)))