一、LeetCode 669 修建二叉搜索树

题目链接：669.修建二叉搜索树https://leetcode.cn/problems/trim-a-binary-search-tree/description/

思路：递归三部曲-定参数、返回值-定终止条件-定单层递归逻辑

class Solution {
    public TreeNode trimBST(TreeNode root, int low, int high) {
        if(root == null){
            return root;
        }
        if(root.val < low){
            //剪枝操作，root的左子树肯定都在区间之外，所以直接返回root的右子树
            return trimBST(root.right,low,high);
        }
        if(root.val > high){
            return trimBST(root.left,low,high);
        }
        //递归处理
        root.left = trimBST(root.left,low,high);
        root.right = trimBST(root.right,low,high);
        return root;
    }
}
/**
 * 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;
 *     }
 * }
 */

 二、LeetCode 108 将有序数组转换为二叉搜索树

题目链接：108.将有序数组转换为二叉搜索树https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/description/

思路：分割数组 左闭右闭 递归造树

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        int n = nums.length;
        int left = 0;
        int right = n-1;
        return travel(nums,0,n-1);

    }
    public TreeNode travel(int[] nums, int left, int right){
        if(left > right){
            return null;
        }
        int index = left + (right-left)/2;
        TreeNode root = new TreeNode(nums[index]);
        //左闭右闭，中间节点已入树
        root.left = travel(nums, left, index-1);
        root.right = travel(nums, index+1, right);
        return root;
    }
}
/**
 * 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;
 *     }
 * }
 */

三、LeetCode 538 把二叉搜索树转换为累加树

题目链接：538.把二叉搜索树转换为累加树https://leetcode.cn/problems/convert-bst-to-greater-tree/

思路：二叉搜索树的中序遍历结果是顺序数组->反中序遍历然后顺序累加可得结果

 

class Solution {
    //记录前一个节点的值
    int pre = 0;
    public TreeNode convertBST(TreeNode root) {
        travel(root);
        return root;
    }
    public void travel(TreeNode root){//按右中左遍历
        if(root == null){
            return;
        }
        travel(root.right);
        //累加求和 保存累加值
        pre += root.val;
        root.val = pre;
        travel(root.left);
    }
}
/**
 * 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;
 *     }
 * }
 */

四、小结

        恨残照，犹有一竿红、怪人催去早 \*AVA*/  $^*^$