99.Recover-Binary-Search-Tree

99. Recover Binary Search Tree

题目地址

https://leetcode.com/problems/recover-binary-search-tree/

题目描述

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?

代码

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

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

  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) {
          // second swap occurence
          x = nums.get(i);
        } 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);
    }
  }

}

Approach #2 Iterative Inorder Traversal

Complexity Analysis

• Time complexity : O(1) in the best case, and O(N) in the worst case when one of the swapped nodes is a rightmost leaf.

• Space complexity : up to O(H) to keep the stack where H is a tree height.

class Solution {
  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);
  }

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

}

Approach #3 Recursive Inorder Traversal

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

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

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

}

Approach #4 Morris Inorder Traversal

class Solution {
    public void recoverTree(TreeNode root) {
    TreeNode x = null, y = null, pred = null, predecessor = null;
    while (root != null) {
      if (root.left !== null) {
        predecessor = root.left;
        while (predecessor.right != null && predecessor.right != root) {
          predecessor = predecessor.right;
        }

        if (predecessor.right == null) {
          predecessor.right = root;
          root = root.left;
        } else {
          if (pred != null && root.val < pred.val) {
            y = root;
            if (x == null) x = pred;
          }
          pred = root;
          predecessor.right = null;
          root = root.right;
        }
      } 
    }

    swap(x , y);
  }


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