129. Sum Root to Leaf Numbers
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Traverse the tree and check for leaf nodes specifically that are left children. Consider using a depth-first search (DFS) approach.
The strategy is to recursively traverse the tree, and at each node, check if it has a left child that is a leaf. If it is, we add its value to the sum. We continue traversing until all nodes are visited.
Approach
- Define a helper function
dfs(node)that will traverse the tree:- Check if the current node is None; if yes, return 0.
- Check if the left child of the node is a leaf (i.e., has no left or right child). If it is, add its value to the sum.
- Recursively call the helper for the left and right children of the current node.
- Return the sum of values from the left and right child calls plus the leaf node value if applicable.
- Call this
dfsfunction starting from the root of the tree.
class Solution:
def sumNumbers(self, root: Optional[TreeNode]) -> int:
def dfs(node):
if not node:
return 0
res = 0
if node.left and not node.left.left and not node.left.right:
res += node.left.val
res += dfs(node.left)
res += dfs(node.right)
return res
return dfs(root)
# 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, sum_so_far):
if root is None:
return 0
res = 0
sum_so_far = sum_so_far * 10 + root.val
if root.left is None and root.right is None:
return sum_so_far
res += dfs(root.left, sum_so_far)
res += dfs(root.right, sum_so_far)
return res
res = dfs(root, 0)
return res