Recover Binary Search Tree

This is No.99 in LeetCode.

Here are approaches to this problem:

Approach 1: Sort an Almost Sorted Array Where Two Elements Are Swapped

Algorithm

Here is the algorithm:

1. Construct inorder traversal of the tree. It should be an almost sorted list where only two elements are swapped.

2. Identify two swapped elements x and y in an almost sorted array in linear time.

3. Traverse the tree again. Change value x to y and value y to x.

class Solution {
  public void inorder(TreeNode root, List<Integer> nums) {
    if (root == null) return;
    inorder(root.left, nums);
    nums.add(root.val);
    inorder(root.right, nums);
  }

  public int[] findTwoSwapped(List<Integer> nums) {
    int n = nums.size();
    int x = -1, y = -1;
    for(int i = 0; i < n - 1; ++i) {
      if (nums.get(i + 1) < nums.get(i)) {
        y = nums.get(i + 1);
        // first swap occurence
        if (x == -1) x = nums.get(i);
        // second swap occurence
        else break;
      }
    }
    return new int[]{x, y};
  }

  public void recover(TreeNode r, int count, int x, int y) {
    if (r != null) {
      if (r.val == x || r.val == y) {
        r.val = r.val == x ? y : x;
        if (--count == 0) return;
      }
      recover(r.left, count, x, y);
      recover(r.right, count, x, y);
    }
  }

  public void recoverTree(TreeNode root) {
    List<Integer> nums = new ArrayList();
    inorder(root, nums);
    int[] swapped = findTwoSwapped(nums);
    recover(root, 2, swapped[0], swapped[1]);
  }
}

Approach 2: Iterative Inorder Traversal

class Solution {
  public void swap(TreeNode a, TreeNode b) {
    int tmp = a.val;
    a.val = b.val;
    b.val = tmp;
  }

  public void recoverTree(TreeNode root) {
    Deque<TreeNode> stack = new ArrayDeque();
    TreeNode x = null, y = null, pred = null;

    while (!stack.isEmpty() || root != null) {
      while (root != null) {
        stack.add(root);
        root = root.left;
      }
      root = stack.removeLast();
      if (pred != null && root.val < pred.val) {
        y = root;
        if (x == null) x = pred;
        else break;
      }
      pred = root;
      root = root.right;
    }

    swap(x, y);
  }
}

Approach 3: Recursive Inorder Traversal

This is recursion version of approach 2.

class Solution {
  TreeNode x = null, y = null, pred = null;

  public void swap(TreeNode a, TreeNode b) {
    int tmp = a.val;
    a.val = b.val;
    b.val = tmp;
  }

  public void findTwoSwapped(TreeNode root) {
    if (root == null) return;
    findTwoSwapped(root.left);
    if (pred != null && root.val < pred.val) {
      y = root;
      if (x == null) x = pred;
      else return;
    }
    pred = root;
    findTwoSwapped(root.right);
  }

  public void recoverTree(TreeNode root) {
    findTwoSwapped(root);
    swap(x, y);
  }
}