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  • 669. Trim a Binary Search Tree

    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

    M1: recursion

    如果root.val比R大,说明要trim掉右边的节点,要返回的subtree在root的左边,即递归到root.left;

    如果root.val比L小,说明要trim掉左边的节点,要返回的subtree在root的右边,递归到root.right;

    如果root.val在[L, R],说明有可能左右两边都需要trim,分别递归左右节点

    time: O(n), space: O(height)

    /**
     * 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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10197446.html
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