Backtracking: 112. Path Sum

112. Path Sum

 

# 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 hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        # DFS로..
        def dfs(node, curSum):
            if not node:
                return False

            curSum += node.val
            # not node.left and not node.right: 왼쪽, 오른쪽 둘다 없으면: 리프 노드임
            
            if not node.left and not node.right:
                # 리프 노드에서 현재까지의 curSum 이 targetSum 와 같으면 True
                return curSum == targetSum

            return (dfs(node.left, curSum) or 
                    dfs(node.right, curSum))

        return dfs(root, 0)

 

 

https://youtu.be/LSKQyOz_P8I?si=C5_7xA-Yqq7ZBMfV

 

이제 풀림

 

# 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 hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        # root에서 leaf 까지 sum 중 targetSum 이 있으면 True, 아니면, false

        def dfs(node, pathsum):
            if not node: # 없으면
                return False
            
            if not node.left and not node.right and ((pathsum + node.val) == targetSum): # 리프노드면
                return True

            if dfs(node.left, pathsum + node.val): # true 면
                return True
            
            if dfs(node.right, pathsum + node.val): # true 면
                return True

            return False

        return dfs(root, 0)