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., $s = s \times 10 + root.val$.
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcsumNumbers(root*TreeNode)int{vardfsfunc(*TreeNode,int)intdfs=func(root*TreeNode,sint)int{ifroot==nil{return0}s=s*10+root.Valifroot.Left==nil&&root.Right==nil{returns}returndfs(root.Left,s)+dfs(root.Right,s)}returndfs(root,0)}