Description#
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.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/0129.Sum%20Root%20to%20Leaf%20Numbers/images/num1tree.jpg)
Input: root = [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:
![](https://spcdn.pages.dev/leetcode/problems/0129.Sum%20Root%20to%20Leaf%20Numbers/images/num2tree.jpg)
Input: root = [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
.
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
.
Solutions#
Solution 1: DFS#
We can design a function $dfs(root, s)$, which represents the sum of all path numbers from the current node $root$ to the leaf nodes, given that the current path number is $s$. The answer is $dfs(root, 0)$.
The calculation of the function $dfs(root, s)$ is as follows:
- If the current node $root$ is null, return $0$.
- Otherwise, add the value of the current node to $s$, i.e., $s = s \times 10 + root.val$.
- If the current node is a leaf node, return $s$.
- Otherwise, return $dfs(root.left, s) + dfs(root.right, s)$.
The time complexity is $O(n)$, and the space complexity is $O(\log n)$. Here, $n$ is the number of nodes in the binary tree.
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| # 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:
def dfs(root, s):
if root is None:
return 0
s = s * 10 + root.val
if root.left is None and root.right is None:
return s
return dfs(root.left, s) + dfs(root.right, s)
return dfs(root, 0)
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| /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int sumNumbers(TreeNode root) {
return dfs(root, 0);
}
private int dfs(TreeNode root, int s) {
if (root == null) {
return 0;
}
s = s * 10 + root.val;
if (root.left == null && root.right == null) {
return s;
}
return dfs(root.left, s) + dfs(root.right, s);
}
}
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| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode* root) {
function<int(TreeNode*, int)> dfs = [&](TreeNode* root, int s) -> int {
if (!root) return 0;
s = s * 10 + root->val;
if (!root->left && !root->right) return s;
return dfs(root->left, s) + dfs(root->right, s);
};
return dfs(root, 0);
}
};
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| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func sumNumbers(root *TreeNode) int {
var dfs func(*TreeNode, int) int
dfs = func(root *TreeNode, s int) int {
if root == nil {
return 0
}
s = s*10 + root.Val
if root.Left == nil && root.Right == nil {
return s
}
return dfs(root.Left, s) + dfs(root.Right, s)
}
return dfs(root, 0)
}
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| /**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function sumNumbers(root: TreeNode | null): number {
function dfs(root: TreeNode | null, s: number): number {
if (!root) return 0;
s = s * 10 + root.val;
if (!root.left && !root.right) return s;
return dfs(root.left, s) + dfs(root.right, s);
}
return dfs(root, 0);
}
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| // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(node: &Option<Rc<RefCell<TreeNode>>>, mut num: i32) -> i32 {
if node.is_none() {
return 0;
}
let node = node.as_ref().unwrap().borrow();
num = num * 10 + node.val;
if node.left.is_none() && node.right.is_none() {
return num;
}
Self::dfs(&node.left, num) + Self::dfs(&node.right, num)
}
pub fn sum_numbers(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
Self::dfs(&root, 0)
}
}
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| /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var sumNumbers = function (root) {
function dfs(root, s) {
if (!root) return 0;
s = s * 10 + root.val;
if (!root.left && !root.right) return s;
return dfs(root.left, s) + dfs(root.right, s);
}
return dfs(root, 0);
};
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| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int dfs(struct TreeNode* root, int num) {
if (!root) {
return 0;
}
num = num * 10 + root->val;
if (!root->left && !root->right) {
return num;
}
return dfs(root->left, num) + dfs(root->right, num);
}
int sumNumbers(struct TreeNode* root) {
return dfs(root, 0);
}
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