99. Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

Output: [3,1,null,null,2]

   3
  /
 1
  \
   2

Example 2:

Input: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

Output: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3

Follow up:

• A solution using O(n) space is pretty straight forward.
• Could you devise a constant space solution?

Code:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def recoverTree(self, root):
        res, stack, first, second = None, [], None, None
        while True:
            while root:
                stack.append(root)
                root = root.left
            if not stack:
                break 
            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.inorder(root)
        self.first.val, self.second.val = self.second.val, self.first.val

    def inorder(self, root):
        if not root:
            return 
        self.inorder(root.left)
        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
        self.inorder(root.right)

    # 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)
            res.append(root)
            self.helper(root.right, res)