530.二叉搜索树的最小绝对差
题目链接
https://leetcode.cn/problems/minimum-absolute-difference-in-bst/description/
题目描述
思路
遇到在二叉搜索树上求什么最值，求差值之类的，都要思考一下二叉搜索树可是有序的，要利用好这一特点。二叉树的中序遍历是从小到大的顺序
1、中序+遍历
先中序遍历得到二叉搜索树的有序数组，再遍历一遍数组，找到最小差值
class Solution {public int getMinimumDifference(TreeNode root) {int min = Integer.MAX_VALUE;List<Integer> list = new ArrayList<>();midorder(root,list);for (int i = 1; i < list.size(); i++) {min = Math.min(min,list.get(i)-list.get(i-1));}return min;}private static void midorder(TreeNode root,List<Integer> list){if(root==null) return;midorder(root.left,list);list.add(root.val);midorder(root.right,list);}
}
2、双指针
class Solution {TreeNode pre;// 记录上一个遍历的结点int result = Integer.MAX_VALUE;public int getMinimumDifference(TreeNode root) {if(root==null)return 0;traversal(root);return result;}public void traversal(TreeNode root){if(root==null)return;//左traversal(root.left);//中if(pre!=null){result = Math.min(result,root.val-pre.val);}pre = root;//右traversal(root.right);}
}
501.二叉搜索树中的众数
题目链接
https://leetcode.cn/problems/find-mode-in-binary-search-tree/description/
题目描述
思路
将每个节点以及节点出现的次数存放到map集合中，遍历集合的值，找到最大的即出现次数最多的，再找到出现次数为最大值的所有键
class Solution {public int[] findMode(TreeNode root) {Map<Integer,Integer> map = new HashMap<>();midorder(root,map);int maxValue = 0;//找出出现次数最多的值for (Integer value : map.values()) {maxValue = Math.max(maxValue,value);}//找到出现次数最多对应的键值//创建一个数组存放出现次数最多的值List<Integer> list = new ArrayList<>();for (Integer integer : map.keySet()) {if(map.get(integer)==maxValue){list.add(integer);}}int[] arr = new int[list.size()];for (int i = 0; i < list.size(); i++) {arr[i] = list.get(i);}return arr;}private void midorder(TreeNode root, Map<Integer,Integer> map){if(root==null) return;midorder(root.left,map);map.put(root.val, map.getOrDefault(root.val,0)+1);midorder(root.right,map);}
}
思路2
findMode1函数中，按照中序遍历去进行判断pre为空（即刚开始）或者pre！=当前节点值（说明已经开始计算下一个节点的出现次数了）的时候，都将count置为1
只有pre==当前节点值的时候，才将count加加；
接下来去判断count和maxCount，刚开始会把一些无关的都添加到列表中，但是在比较的过程中会不断的更新maxCount，所以需要清空列表去添加最大的，最后在主函数中调用这个函数，再用数组去存储
class Solution {ArrayList<Integer> resList;int maxCount;int count;TreeNode pre;public int[] findMode(TreeNode root) {resList = new ArrayList<>();maxCount = 0;count = 0;pre = null;findMode1(root);int[] res = new int[resList.size()];for (int i = 0; i < resList.size(); i++) {res[i] = resList.get(i);}return res;}public void findMode1(TreeNode root) {if (root == null) {return;}findMode1(root.left);int rootValue = root.val;// 计数if (pre == null || rootValue != pre.val) {count = 1;} else {count++;}// 更新结果以及maxCountif (count > maxCount) {resList.clear();resList.add(rootValue);maxCount = count;} else if (count == maxCount) {resList.add(rootValue);}pre = root;findMode1(root.right);}
}
236. 二叉树的最近公共祖先
题目链接
https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/
题目描述
思路
没太明白
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root == null || root == p || root == q) { // 递归结束条件return root;}// 后序遍历TreeNode left = lowestCommonAncestor(root.left, p, q);TreeNode right = lowestCommonAncestor(root.right, p, q);if(left == null && right == null) { // 若未找到节点 p 或 qreturn null;}else if(left == null && right != null) { // 若找到一个节点return right;}else if(left != null && right == null) { // 若找到一个节点return left;}else { // 若找到两个节点return root;}}
}
