AC Python 88ms simple recursion


  • 0
    def _toBST(self, nums, l, r):
        if l == r:
            return None
        m = (l + r) >> 1
        root = TreeNode(nums[m])
        root.left = self._toBST(nums, l, m)
        root.right = self._toBST(nums, m + 1, r)
        return root
    
    def sortedArrayToBST(self, nums):
        return self._toBST(nums, 0, len(nums))
    
    
    # 32 / 32 test cases passed.
    # Status: Accepted
    # Runtime: 88 ms
    # 99.30%
    

    The logic is almost the same as a post order traversal of a binary tree. We need to build the left and right subtree before we can connect them to the root.


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