Shulou(Shulou.com)06/01 Report--

This article introduces the relevant knowledge of "Java binary search tree and array search method". In the operation of actual cases, many people will encounter such a dilemma, so let the editor lead you to learn how to deal with these situations. I hope you can read it carefully and be able to achieve something!

Topic one

Solution method

/ * 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 ans Int pre; public int minDiffInBST (TreeNode root) {ans = Integer.MAX_VALUE; pre=-1; method (root); return ans;} public void method (TreeNode root) {if (root==null) return; method (root.left); if (pre==-1) {pre= root.val } else {ans = Math.min (ans,root.val-pre); pre = root.val;} method (root.right);}} topic 2

Solution method

Class Solution {public int dominantIndex (int [] nums) {int f = Integer.MIN_VALUE; int fi = 0; int s = Integer.MIN_VALUE; int si = 0; for (int I = 0; if) {s = f; f = nums [I]; fi = I } else if (Nums [I] > s) {s = nums [I];}} if (nums.length==1) return 0; if (2s)

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