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 = 2Output:     1      \       2

 

Example 2:

Input:     3   / \  0   4   \    2   /  1  L = 1  R = 3Output:       3     /    2     / 1 思路：考的是二叉树节点的删除，只要满足条件的删除就好了，得注意的是一个节点是有两个孩子还是小于两孩子；

/** * 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)                                       //如果一开始传进来的是空，则返回空；            return null;        root.left = trimBST(root.left,L,R);           //递归的删除，从底层删起        root.right = trimBST(root.right,L,R);        if (root.val
     
      R){                          //这里是如何删除满足条件的节点的代码            if (root.left!=null&&root.right!=null){                 root.val = findMax(root.left).val;            //这是有两个孩子的情况                trimBST(root.left,root.val-1,root.val+1);            }            else {                                            //只有一个孩子或无孩子                if (root.left==null)                    root = root.right;                else if (root.right==null)                    root = root.left;            }        }        return root;    }    public TreeNode findMax(TreeNode root){        if (root==null)            return null;        if (root.right==null)            return root;        else             return findMax(root.right);    }}