[Leetcode]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

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<L||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);
    }
}