230: Time 97.61% and Space 83.27%, Solution with step by step explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nThe problem asks us to find the kth smallest value in a binary search tre
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nThe problem asks us to find the kth smallest value in a binary search tree. One way to do this is to traverse the tree in-order (left, root, right) and keep track of the number of nodes we have visited. Once we have visited k nodes, the next node we visit will be the kth smallest node.\n\nWe can do this using a stack to store the nodes we have visited so far. We start at the root of the tree and traverse as far left as possible, adding each node to the stack. Once we reach the leftmost node, we pop it off the stack (which is the smallest node we have seen so far) and decrement k. If k is 0, we have found the kth smallest value and can return it.\n\nIf k is not 0, we move to the right child of the node we just processed and repeat the process. This will visit the nodes in ascending order of value until we find the kth smallest node.\n\nNote that this algorithm has a time complexity of O(n) in the worst case (when the tree is skewed), since we need to visit every node in the tree. However, in the average case (when the tree is balanced), it has a time complexity of O(log n) since we only need to visit a subset of the nodes. Additionally, the space complexity is O(h) where h is the height of the tree, since we need to store at most h nodes on the stack at any given time.\n\n# Complexity\n- Time complexity:\n97.61%\n\n- Space complexity:\n83.27%\n\n# Code\n```\nclass Solution:\n    def kthSmallest(self, root: TreeNode, k: int) -> int:\n        # create a stack to store the nodes\n        stack = []\n        # start at the root of the tree\n        current = root\n        \n        # loop until we have processed all nodes and found the kth smallest value\n        while True:\n            # traverse as far left as possible from the current node, adding each node to the stack\n            while current is not None:\n                stack.append(current)\n                current = current.left\n            \n            # if the stack is empty, we have processed all nodes and can exit the loop\n            if not stack:\n                break\n                \n            # pop the top node off the stack (which is the next smallest node) and decrement k\n            node = stack.pop()\n            k -= 1\n            \n            # if k is 0, we have found the kth smallest value and can return it\n            if k == 0:\n                return node.val\n            \n            # set the current node to the right child of the node we just processed\n            current = node.right\n\n```
['Tree', 'Depth-First Search', 'Binary Search Tree', 'Python', 'Python3']