Python Simple O(n) DFS

  • 2
    def findTarget(self, root, k):
        a = set()
        self.f = False
        def dfs(root, k):
            if not root:
            if root.val not in a:
                a.add(k - root.val)
                self.f = True
            dfs(root.left, k)
            dfs(root.right, k)
        dfs(root, k)
        return self.f

  • 2

    Since it's a BST, a better solution would take advantage of it by using two pointers pointing at the smallest and the largest.

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