669.Trim-a-Binary-Search-Tree

669. Trim a Binary Search Tree

题目地址

https://leetcode.com/problems/trim-a-binary-search-tree/

题目描述

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

代码

Approach #1 Recursion

Time Complexity:O(N) && Space Complexity: O(N)

/**
 * 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;
      if (root.val > R) return trimBST(root.left, L, R);
      if (root.val < L) return trimBST(root.right, L, R);

      root.left = trimBST(root.left, L, R);
      root.right = trimBST(root.right, L, R);
      return root;
    }
}

Approach #2 Iteration

class Solution {
    public TreeNode trimBST(TreeNode root, int L, int R) {
        if (root == null) {
            return root;
        }
        // 1. Find a valid root which is used to return.
        while (root.val < L || root.val > R) {
            if (root.val < L) {
                root = root.right;
            }
            if (root.val > R) {
                root = root.left;
            }
        }
        TreeNode dummy = root;
        // 2. Remove the invalid nodes from left subtree.
        while (dummy != null) {
            while (dummy.left != null && dummy.left.val < L) {
                dummy.left = dummy.left.right; 
                // If the left child is smaller than L, then we just keep the right subtree of it. 
            }
            dummy = dummy.left;
        }
        dummy = root;
        // 3. Remove the invalid nodes from right subtree
        while (dummy != null) {
            while (dummy.right != null && dummy.right.val > R) {
                dummy.right = dummy.right.left;
                // If the right child is biggrt than R, then we just keep the left subtree of it. 
            }
            dummy = dummy.right;
        }
        return root;
    }
}