From fdc24da4fff03786577b15afeeb7b4c8e55f2bb1 Mon Sep 17 00:00:00 2001
From: Sangeeth Sudheer <sangeeth96@users.noreply.github.com>
Date: Wed, 20 Apr 2022 05:20:11 +0530
Subject: [PATCH] add py3 soln

---
 0099_recover-binary-search-tree/README.md     | 26 +++++
 .../python3/linear_space.py                   | 94 +++++++++++++++++++
 .../python3/solution.py                       |  0
 3 files changed, 120 insertions(+)
 create mode 100644 0099_recover-binary-search-tree/README.md
 create mode 100644 0099_recover-binary-search-tree/python3/linear_space.py
 create mode 100644 0099_recover-binary-search-tree/python3/solution.py

diff --git a/0099_recover-binary-search-tree/README.md b/0099_recover-binary-search-tree/README.md
new file mode 100644
index 0000000..bb6e1b4
--- /dev/null
+++ b/0099_recover-binary-search-tree/README.md
@@ -0,0 +1,26 @@
+You are given the `root` of a binary search tree (BST), where the values of **exactly** two nodes of the tree were swapped by mistake. _Recover the tree without changing its structure_.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/10/28/recover1.jpg)
+
+    Input: root = [1,3,null,null,2]
+    Output: [3,1,null,null,2]
+    Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.
+    
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2020/10/28/recover2.jpg)
+
+    Input: root = [3,1,4,null,null,2]
+    Output: [2,1,4,null,null,3]
+    Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.
+    
+
+**Constraints:**
+
+*   The number of nodes in the tree is in the range `[2, 1000]`.
+*   `-231 <= Node.val <= 231 - 1`
+
+**Follow up:** A solution using `O(n)` space is pretty straight-forward. Could you devise a constant `O(1)` space solution?
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diff --git a/0099_recover-binary-search-tree/python3/linear_space.py b/0099_recover-binary-search-tree/python3/linear_space.py
new file mode 100644
index 0000000..5c0e7cb
--- /dev/null
+++ b/0099_recover-binary-search-tree/python3/linear_space.py
@@ -0,0 +1,94 @@
+# Time: O(N)
+# Space: O(N)
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    items = swaps = None
+    
+    def recoverTree(self, root: Optional[TreeNode]) -> None:
+        """
+        Do not return anything, modify root in-place instead.
+        """
+        # Get the almost-sorted tree as an array
+        self.items = self.inorder(root)
+        
+        # Find the elements that need to be swapped
+        self.swaps = self.identify_swaps(self.items)
+        
+        # Walk the tree and fix the matching values
+        self.walk_fix(root, 0)
+        
+    
+    def inorder(self, node):
+        """
+        Return list of elements in almost sorted order (BST property
+        implies that in-order traversal produces sorted list)
+        """
+        
+        return self.inorder(node.left) + [node.val] + self.inorder(node.right) if node is not None else []
+        
+        
+    def identify_swaps(self, items):
+        # We need to swap two values so keep two variables
+        x = y = None
+
+        for i in range(len(items) - 1):
+            # Problem states that we at most need to swap one item
+            # at index i to another item at index j to get the sorted
+            # list. If this is the case, the larger value to swap will
+            # be seen first and later on, the smaller value will be seen
+            
+            # e.g. 1 4 3 2
+            # and  2 1
+            if items[i] < items[i + 1]:
+                pass
+            else:
+                # Condition fails when we see (4, 3) in first e.g and (2, 1)
+                # in second example. We immediately assign i+1th value to cover
+                # a case like the second example where swaps are right next to each
+                # in last iteration
+                y = items[i + 1]
+
+                # In (2, 1)'s case, immediately goes to this block and iteration stops
+                # when iteration condition becomes False.
+                #
+                # In first example's case, we get to (4, 3) in which case:
+                #
+                # y = items[i + 1] = 3
+                # x = items[i] = 4
+                #
+                # We keep looping until we hit (3, 2) and this time, it's the lower value
+                # so y is reassigned to 2. x is already set previously to 4 so now we can
+                # break and return the values to swap.
+                if x is None:
+                    x = items[i]
+                else:
+                    break
+
+        return {
+            x: y,
+            y: x
+        }
+        
+        
+    def walk_fix(self, node, i):
+        """
+        Does in-order traversal of BST and if node.val matches
+        a key in self.swaps, replaces it with the swapped value
+        """
+        if node is None: return
+
+        self.walk_fix(node.left, i)
+
+        if node.val in self.swaps:
+            node.val = self.swaps[node.val]
+
+        i += 1
+
+        self.walk_fix(node.right, i)
+    
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diff --git a/0099_recover-binary-search-tree/python3/solution.py b/0099_recover-binary-search-tree/python3/solution.py
new file mode 100644
index 0000000..e69de29