669. Trim a Binary Search Tree.md

669. Trim a Binary Search Tree

Question:

Given a binary search tree and the lowest and highest boundaries as L and R, trim the tree so that all its elements lies in [L, R] (R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.

Example 1:

Input: 
    1
   / \
  0   2

  L = 1
  R = 2

Output: 
    1
      \
       2

Example 2:

Input: 
    3
   / \
  0   4
   \
    2
   /
  1

  L = 1
  R = 3

Output: 
      3
     / 
   2   
  /
 1

Solution:

• Method 1: divide and conquer (O(V + E), time complexity of dfs )
  • divide: we do the trim for left and right node.
  • if current node is within the range, we just return current node.
  • if current node needs to be trimmed, we need to append the right node to the very right of left subtree and return the left node of current node.

  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  class Solution {
      public TreeNode trimBST(TreeNode root, int L, int R) {
          if(root == null) return root;
          TreeNode left = trimBST(root.left, L, R);
          root.left = left;
          TreeNode right = trimBST(root.right, L, R);
          root.right = right;
          if(root.val < L || root.val > R){
              if(root.left == null) return root.right;
              else{
                  TreeNode node = root.left;
                  while(node.right != null){
                      node = node.right;
                  }
                  node.right = root.right;
                  root.right = null;
                  return root.left;
              }
          }
          return root;
      }
  }