Python iterative solution (average O(lgn) space, one pass).

  • 5

    Based on the idea here (, this is an implementation iteratively. The other two implementations can serve for comparison:

    # average O(lgn) space (worst case O(n) space), iteratively, one-pass
    def recoverTree(self, root):
        res, stack, first, second = None, [], None, None
        while True:
            while root:
                root = root.left
            if not stack:
            node = stack.pop()
            # first time occurs reversed order
            if res and res.val > node.val:
                if not first:
                     first = res
                # first or second time occurs reversed order
                second = node
            res = node
            root = node.right
        first.val, second.val = second.val, first.val
    # average O(lgn) space (worst case, O(n) space), recursively, one-pass 
    def recoverTree2(self, root):
        self.prevNode = TreeNode(-sys.maxsize-1)
        self.first, self.second = None, None
        self.first.val, self.second.val = self.second.val, self.first.val
    def inorder(self, root):
        if not root:
        if not self.first and self.prevNode.val > root.val:
            self.first, self.second = self.prevNode, root
        if self.first and self.prevNode.val > root.val:
            self.second = root
        self.prevNode = root
    # average O(n+lgn) space, worst case O(2n) space, recursively, two-pass
    def recoverTree3(self, root):
        res = []
        self.helper(root, res)
        first, second = None, None
        for i in xrange(1, len(res)):
            if not first and res[i-1].val > res[i].val:
                first, second = res[i-1], res[i]
            if first and res[i-1].val > res[i].val:
                second = res[i]
        first.val, second.val = second.val, first.val
    def helper(self, root, res):
        if root:
            self.helper(root.left, res)
            self.helper(root.right, res)

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