思路： 这道题和删除节点异曲同工。不过要注意避坑：当遍历到不在范围内的节点时，不要直接返回null或直接返回其左或右孩子，而是继续对其左或右孩子做递归。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public TreeNode trimBST(TreeNode root, int low, int high) {return dfs(root,low,high);} public TreeNode dfs(TreeNode root, int low, int high){if(root==null) return null;if(root.val<low) return dfs(root.right,low,high);//注意，不要直接返回root.rightif(root.val>high) return dfs(root.left,low,high);root.left=dfs(root.left,low,high);root.right=dfs(root.right,low,high);return root;}
}

思路： 这道题是使用二分查找的思路递归构建二叉树。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public TreeNode sortedArrayToBST(int[] nums) {return getSortedArrayToBST(nums,0,nums.length-1);}public TreeNode getSortedArrayToBST(int[] nums,int left,int right){if(left>right) return null;if(left==right) return new TreeNode(nums[left]);int mid=left+(right-left)/2;TreeNode root=new TreeNode(nums[mid]);root.left=getSortedArrayToBST(nums,left,mid-1);root.right=getSortedArrayToBST(nums,mid+1,right);return root;}
}

思路： 使用双指针的方法，右中左的顺序遍历数组。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {TreeNode pre;public TreeNode convertBST(TreeNode root) {getConvertBST(root);return root;}public void getConvertBST(TreeNode node){if(node==null) return;getConvertBST(node.right);if(pre!=null){node.val=node.val+pre.val;}pre=node;getConvertBST(node.left);}
}