Problem 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:

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9
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:

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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.

Solution:

A simple DFS solution, use an integer to store the current sum. For every node, update the val as val*10 plus node’s data.

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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, curNum):
            # print("=====")
            if not root.left and not root.right:
                return curNum*10+root.val
            left, right = 0, 0
            if root.left:
                left = dfs(root.left, curNum*10 + root.val)
            if root.right:
                right = dfs(root.right, curNum*10 + root.val)
            return left + right
        return dfs(root, 0)
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/**
 * 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) {
        if (!root) return 0;
        return helper(root, 0);
    }
    int helper(TreeNode* root, int x) {
        if (!root->right && !root->left)
            return 10 * x + root->val;
        int val = 0;
        if (root->left)
            val += helper(root->left, 10 * x + root->val);
        if (root->right)
            val += helper(root->right, 10 * x + root->val);
        return val;
    }
};

BFS

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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:
        # BFS
        queue = deque([(root, root.val)])
        res = 0
        while queue:
            front, curSum = queue.popleft()
            if not front.left and not front.right:
                print(curSum)
                res += curSum
                continue
            
            if front.left:
                queue.append((front.left, curSum*10+front.left.val))
            if front.right:
                queue.append((front.right, curSum*10+front.right.val))
        return res

time complexity: $O(n)$
space complexity: $O(logn)$
reference: