您好,欢迎来到意榕旅游网。
搜索
您的当前位置:首页代码随想录day23

代码随想录day23

来源:意榕旅游网

一、修剪二叉搜索树(LeetCode669)

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) {
        if(root == null)    return null;
        if(root.val < low){
            TreeNode right = trimBST(root.right,low,high);
            return right;
        }
        if(root.val > high){
            TreeNode left = trimBST(root.left,low,high);
            return left;
        }
        root.left = trimBST(root.left,low,high);
        root.right = trimBST(root.right,low,high);
        return root;
    }
}

二、将有序数组转换为二叉搜索树(LeetCode108)

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 sortedArrayToBST(int[] nums) {
        return travesal(nums,0,nums.length-1);
    }
    TreeNode travesal(int[] nums,int left,int right){
        if(left > right){
            return null;
        }
        // int mid = (left + right) / 2; //数组可能会越界
        int mid = left + ((right - left) >> 1);//如果数组长度为偶数,中间位置有两个元素,取靠左边的,另一个就是避免数组越界
        TreeNode root = new TreeNode(nums[mid]);
        root.left = travesal(nums,left,mid-1);
        root.right = travesal(nums,mid+1,right);
        return root;
    }
}

三、把二叉搜索树转换为累加树(LeetCode538)

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 {
    int pre = 0;
    public TreeNode convertBST(TreeNode root) {
        travesal(root);
        return root; 
    }
    void travesal(TreeNode cur){
        if(cur == null) return ;
        travesal(cur.right);
        cur.val += pre;
        pre = cur.val;
        travesal(cur.left);
    }
}

因篇幅问题不能全部显示,请点此查看更多更全内容

Copyright © 2019- yrrf.cn 版权所有 赣ICP备2024042794号-2

违法及侵权请联系:TEL:199 1889 7713 E-MAIL:2724546146@qq.com

本站由北京市万商天勤律师事务所王兴未律师提供法律服务