129. Sum Root to Leaf Numbers
Difficulty: Medium
Topics: Tree
, Depth-First Search
, Binary Tree
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 number123
.
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
.
Solution:
We can use a Depth-First Search (DFS) approach. The idea is to traverse the tree, keeping track of the number formed by the path from the root to the current node. When you reach a leaf node, you add the current number to the total sum.
Let's implement this solution in PHP: 129. Sum Root to Leaf Numbers
Explanation:
TreeNode Class: This class represents each node of the binary tree. Each node has a value (
val
), a left child (left
), and a right child (right
).sumNumbers Function: This is the main function that initiates the DFS traversal by calling the
sumNumbersRecu
function with the root node and an initialcurrentSum
of 0.-
sumNumbersRecu Function:
-
Base Case: If the node is
null
, return 0 (this handles the case where there is no child node). -
Update Current Sum: For each node, update the
currentSum
by multiplying the previouscurrentSum
by 10 and adding the node's value. -
Leaf Node Check: If the node is a leaf (i.e., it has no left or right child), return the
currentSum
as this represents the number formed by the path from the root to this leaf. -
Recursive Calls: If the node is not a leaf, recursively call
sumNumbersRecu
on both the left and right children and return their sum.
-
Base Case: If the node is
Edge Cases:
- The function handles cases where the tree has only one node.
- It correctly computes the sum even for deeper trees, as the depth of the tree will not exceed 10 (as per the problem constraints).
This solution runs efficiently with a time complexity of O(n), where n is the number of nodes in the tree. This is because each node is visited exactly once. The space complexity is O(h), where h is the height of the tree, due to the recursion stack.
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