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129. Sum Root to Leaf Numbers

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:

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:

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.

  • /**
 * 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);
    }
}

  • /**
 * 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);
    }
};

  • # 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)


  • /**
 * 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)
}

  • /**
 * 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);
}


  • /**
 * 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);
};


  • // 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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