669. 修剪二叉搜索树
 

/**
 * 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 trimBST(TreeNode root, int low, int high) {
        if(root == null){
            return null;
        }
        //比low小，修剪左枝,找右枝符合范围的结点
        if(root.val < low ){
            return trimBST(root.right,low,high);
        }
        //比high大，修建右枝，找左枝符合范围的结点
        if(root.val > high){
            return trimBST(root.left,low,high);
        }
        root.left = trimBST(root.left,low,high);
        root.right = trimBST(root.right,low,high);
        return root;
    }
}

这道题感觉理解时候有点抽象（回顾时多思考一下），遇到在合理范围内的结点，则进行保留(return root;),比low小，这个结点用其右孩子代替；比high大，这个结点用其左孩子代替，一直递归。

108. 将有序数组转换为二叉搜索树

/**
 * 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 sortedArrayToBST(nums, 0 , nums.length-1);
    }
    //不断选取区间中间的那个数构造结点
    public TreeNode sortedArrayToBST(int[] nums ,int left ,int right){
        if(right - left < 0 ) return null;
        int mid = (left + right) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = sortedArrayToBST(nums,left,mid-1);
        root.right = sortedArrayToBST(nums,mid+1,right);
        return root;
    }
}

538. 把二叉搜索树转换为累加树

/**
 * 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 {
    //新值等于原树中大于等于其结点的值的和
    //右中左遍历
    int sum = 0;
    public TreeNode convertBST(TreeNode root) {
        //测试用例中有root为空的树
        if(root == null) return null;
        if(root.right !=null){
            convertBST(root.right);
        }
        //ERROR: 此处不能左孩子不能活得其父节点的值
        // int increase = root.right == null ? 0 :root.right.val;
        // root.val += increase;
        sum += root.val;
        root.val = sum ;
        if(root.left != null){
            convertBST(root.left);
        }
        return root;
    }
}