代码随想录算法训练营第二十三天 | 669. 修剪二叉搜索树|108.将有序数组转换为二叉搜索树|538.把二叉搜索树转换为累加树

669. 修剪二叉搜索树

题解及想法

使用中序递归法

当root的元素小于low的数值，那么应该递归右子树，并返回右子树符合条件的头结点。当root的元素大于high的，那么应该递归左子树，并返回左子树符合条件的头结点。接下来要将下一层处理完左子树的结果赋给root->left，处理完右子树的结果赋给root->right。最后返回root节点

/**
 * 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;
        //中
        if(root.val < low){ 
            return trimBST(root.right,low,high);
        }
        if(root.val > high){
            return trimBST(root.left,low,high);
        }
        //左
        root.left = trimBST(root.left,low,high);
        //右
        root.right = trimBST(root.right,low,high);
        return root;
    }
}

108.将有序数组转换为二叉搜索树

题解及想法

递归法，左闭右闭区间[left, right]。

/**
 * 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 traversal(nums, 0, nums.length - 1);
    }


    private TreeNode traversal(int[] nums, int left, int right) {
		if (left > right) return null;
		int mid = left + ((right - left) >> 1);
		TreeNode root = new TreeNode(nums[mid]);
		root.left = traversal(nums, left, mid - 1);
		root.right = traversal(nums, mid + 1, right);
		return root;
	}
}

538.把二叉搜索树转换为累加树

题解及想法

递归 右中左 根据二叉树搜索的特性，右中左就是从大到小遍历，通过pre指向遍历的节点的前一个，遍历到那个节点就加上pre，再把pre指向当前节点进行递归，就达到累加的目的。

/**
 * 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 {
    int pre = 0; //递归的前一个节点
    public TreeNode convertBST(TreeNode root) {
        traversal(root);
        return root;
    }
    //右中左
    public void traversal(TreeNode root) {
        if(root == null) return;
        traversal(root.right);  //右

        root.val += pre;  //加上前一个大节点的值  中
        pre = root.val;

        traversal(root.left);  //左   
    }
}