代码随想录算法训练营 DAY20 | 二叉树(7)

一、LeetCode 530 二叉搜索树的最小绝对值

题目链接：530.二叉搜索树的最小绝对值https://leetcode.cn/problems/minimum-absolute-difference-in-bst/

思路一：利用搜索二叉树的中序遍历结果为有序数组的性质，将遍历结果保存到数组中，再找最小绝对值。

class Solution {
    List list = new LinkedList<>();
    public int getMinimumDifference(TreeNode root) {
        //二叉搜索树的中序遍历结果是一个有序数组
        midgo(root);
        int ans = Integer.MAX_VALUE;
        for(int i = 1; i < list.size(); i++){
            int temp = list.get(i) - list.get(i-1);
            if( temp < ans){
                ans = temp;
            }
        }
        return ans;
    }
    public void midgo(TreeNode root){
        if(root == null){
            return;
        }
        midgo(root.left);
        list.add(root.val);
        midgo(root.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;
 *     }
 * }
 */

 思路二：利用pre节点记录上个遍历到的节点数值，直接完成递归遍历和计算。

class Solution {
    int result = Integer.MAX_VALUE;
    TreeNode pre = null;
    public int getMinimumDifference(TreeNode root) {
        travel(root);
        return result;
    }
    public void travel(TreeNode root){
        if(root == null){
            return;
        }
        travel(root.left);
        //前一个节点不为空，比较差值与已保存的最小绝对值
        if(pre != null){
            result = Math.min(result,root.val-pre.val);
        }
        //记录前一个节点
        pre = root;
        travel(root.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;
 *     }
 * }
 */

 二、LeetCode 501 二叉搜索树中的众数

题目链接：501.二叉搜索树中的众数https://leetcode.cn/problems/find-mode-in-binary-search-tree/

思路：利用二叉搜索树中序遍历为有序数组的性质，设置pre记录上一个节点，对众数进行计数及结果更新。

 //二叉搜索树特性：中序遍历结果为有序数组
class Solution {
    List list;
    int maxCount;
    int count;
    TreeNode pre;
    public int[] findMode(TreeNode root) {
        list = new ArrayList<>();
        maxCount = 0;
        count = 0;
        pre = null;
        travel(root);
        int n = list.size();
        int[] ans = new int[n];
        for(int i = 0; i < n; i++){
            ans[i] = list.get(i);
        }
        return ans;
    }
    public void travel(TreeNode root){
        if(root == null){
            return;
        }
        travel(root.left);
        //利用中序遍历特性计数
        if(pre == null || root.val != pre.val){
            count = 1;
        }else{
            count++;
        }
        //更新结果及maxCount
        if(count > maxCount){
            maxCount = count;
            //最大值改变，清空结果
            list.clear();
            list.add(root.val);
        }else if(count == maxCount){
            //最大值未变，添加结果
            list.add(root.val);
        }
        pre = root;
        travel(root.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;
 *     }
 * }
 */