目录

1. 二叉树的最大深度  🌟

2. 二叉树的层序遍历  🌟🌟

3. 最短回文串  🌟🌟🌟

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1. 二叉树的最大深度

给定一个二叉树，找出其最大深度。

二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。

说明: 叶子节点是指没有子节点的节点。

示例：
给定二叉树 [3,9,20,null,null,15,7]，

     3/ \\9  20/  \\
15   7

返回它的最大深度 3 。

出处：

https://edu.csdn.net/practice/25949981

代码：

import java.util.*;
public class maxDepth {public final static int NULL = Integer.MIN_VALUE; //用NULL来表示空节点public static class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}}public static class Solution {public int maxDepth(TreeNode root) {if (root == null) {return 0;}int deep = 1;if (root.left == null && root.right == null) {return 1;}int leftDeep = 0;if (root.left != null) {leftDeep = 1 + maxDepth(root.left);}int rightDeep = 0;if (root.right != null) {rightDeep = 1 + maxDepth(root.right);}return deep + leftDeep > rightDeep ? leftDeep : rightDeep;}}public static TreeNode createBinaryTree(Vector<Integer> vec) {if (vec == null || vec.size() == 0) {return null;}Queue<TreeNode> queue = new LinkedList<>();TreeNode root = new TreeNode(vec.get(0));queue.offer(root);int i = 1;while (!queue.isEmpty()) {int size = queue.size();for (int j = 0; j < size; j++) {TreeNode node = queue.poll();if (i < vec.size() && vec.get(i) != NULL) {node.left = new TreeNode(vec.get(i));queue.offer(node.left);}i++;if (i < vec.size() && vec.get(i) != NULL) {node.right = new TreeNode(vec.get(i));queue.offer(node.right);}i++;}}return root;}public static void main(String[] args) {Solution s = new Solution();Integer[] nums = {3,9,20,NULL,NULL,15,7};Vector<Integer> vec = new Vector<Integer>(Arrays.asList(nums));TreeNode root = createBinaryTree(vec);System.out.println(s.maxDepth(root));}
}

输出：

3

2. 二叉树的层序遍历

给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。

示例：
二叉树：[3,9,20,null,null,15,7],

     3/ \\9  20/  \\
15   7

返回其层序遍历结果：

[
[3],
[9,20],
[15,7]
]

出处：

https://edu.csdn.net/practice/25855085

代码：

import java.util.*;
public class levelOrder {public final static int NULL = Integer.MIN_VALUE; //用NULL来表示空节点public static class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}}public static class Solution {public List<List<Integer>> levelOrder(TreeNode root) {List<List<Integer>> l = new ArrayList<>();Queue<TreeNode> q = new LinkedList<TreeNode>();if (root != null) {q.add(root);}while (!q.isEmpty()) {List<Integer> l2 = new ArrayList<>();int number = q.size();while (number > 0) {TreeNode t = q.poll();l2.add(t.val);if (t.left != null) {q.add(t.left);}if (t.right != null) {q.add(t.right);}number--;}l.add(l2);}return l;}}public static TreeNode createBinaryTree(Vector<Integer> vec) {if (vec == null || vec.size() == 0) {return null;}Queue<TreeNode> queue = new LinkedList<>();TreeNode root = new TreeNode(vec.get(0));queue.offer(root);int i = 1;while (!queue.isEmpty()) {int size = queue.size();for (int j = 0; j < size; j++) {TreeNode node = queue.poll();if (i < vec.size() && vec.get(i) != NULL) {node.left = new TreeNode(vec.get(i));queue.offer(node.left);}i++;if (i < vec.size() && vec.get(i) != NULL) {node.right = new TreeNode(vec.get(i));queue.offer(node.right);}i++;}}return root;}public static void main(String[] args) {Solution s = new Solution();Integer[] nums = {3,9,20,NULL,NULL,15,7};Vector<Integer> vec = new Vector<Integer>(Arrays.asList(nums));TreeNode root = createBinaryTree(vec);System.out.println(s.levelOrder(root));}
}

输出：

[[3], [9, 20], [15, 7]]

3. 最短回文串

给定一个字符串 s，你可以通过在字符串前面添加字符将其转换为回文串。找到并返回可以用这种方式转换的最短回文串。

示例 1：

输入：s = "aacecaaa"
输出："aaacecaaa"

示例 2：

输入：s = "abcd"
输出："dcbabcd"

提示：

• 0 <= s.length <= 5 * 10^4
• s 仅由小写英文字母组成

出处：

https://edu.csdn.net/practice/25949983

代码：

import java.util.*;
public class shortestPalindrome {public static class Solution {public static String shortestPalindrome(String s) {StringBuilder r = new StringBuilder(s).reverse();String str = s + "#" + r;int[] next = next(str);String prefix = r.substring(0, r.length() - next[str.length()]);return prefix + s;}private static int[] next(String P) {int[] next = new int[P.length() + 1];next[0] = -1;int k = -1;int i = 1;while (i < next.length) {if (k == -1 || P.charAt(k) == P.charAt(i - 1)) {next[i++] = ++k;} else {k = next[k];}}return next;}}public static void main(String[] args) {Solution s = new Solution();String str = "aacecaaa";System.out.println(s.shortestPalindrome(str));str = "abcd";System.out.println(s.shortestPalindrome(str));}
}

输出：

aaacecaaa
dcbabcd

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