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‎2018.11.29-leetcode443/kiritocly.md

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private static int solution(char[] charArray) {
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if (charArray.length == 0) {
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return 0;
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}
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//遍历整个字符数组
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//压缩后的字符长度
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int resultCount = 0;
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//当前重复字符计数
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int count = 1;
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for (int i = 0; i < charArray.length; i++) {
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//如果遇到某个字符与当前字符不相等或者已经遍历到最后一个字符
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if (i + 1 == charArray.length || charArray[i] != charArray[i + 1]) {
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//记录一下当前字符，并且resultCount加1，比如a,a,b,b,b，扫面到第二个a时
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//a!=b,如果a的个数大于1，先将a后面所有的a替换为数字，如a,a替换为a,2
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charArray[resultCount++] = charArray[i];
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if (count > 1) {
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String temp = String.valueOf(count);
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for (int k = 0; k < temp.length(); k++) {
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charArray[resultCount++] = temp.charAt(k);
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}
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}
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//重新统计下一个字符
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count = 1;
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} else {
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//统计重复的字符个数
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count++;
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}
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}
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return resultCount;
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}

‎2018.12.04-leetcode101/Tony the Cyclist.md

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递归
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```java
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class Solution {
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public boolean isSymmetric(TreeNode root) {
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if (root == null){
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return true;
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}
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else {
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if (check(root.left, root.right)){
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return true;
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}
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else {
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return false;
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}
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}
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}
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public static boolean check(TreeNode node1, TreeNode node2){
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if (node1 == null && node2 == null){
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return true;
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}
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else if (node1 == null || node2 == null){
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return false;
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}
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return node1.val == node2.val && check(node1.right, node2.left) && check(node1.left, node2.right);
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}
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}
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```
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非递归
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```java
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import java.util.*;
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class Solution {
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public boolean isSymmetric(TreeNode root) {
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if ((root == null) || (root.left == null && root.right == null)){
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return true;
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}
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else if (root.left == null || root.right == null){
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return false;
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}
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else {
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ArrayDeque<TreeNode> d = new ArrayDeque<TreeNode>();
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d.addFirst(root.left);
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d.addLast(root.right);
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while (d.size() != 0){
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TreeNode leftnode = d.pollFirst();
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TreeNode rightnode = d.pollLast();
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if (leftnode == null && rightnode == null){
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return true;
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}
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else if (leftnode == null || rightnode == null){
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return false;
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}
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else if (leftnode.val != rightnode.val){
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return false;
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}
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else {
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if (rightnode.right != null && leftnode.left != null){
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d.addLast(rightnode.right);
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d.addFirst(leftnode.left);
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}
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else if (rightnode.right == null && leftnode.left == null){
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}
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else {
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return false;
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}
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if (rightnode.left != null && leftnode.right != null){
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d.addFirst(rightnode.left);
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d.addLast(leftnode.right);
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}
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else if (rightnode.left == null && leftnode.right == null){
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}
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else {
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return false;
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}
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}
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}
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return true;
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}
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}
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}
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```

‎2018.12.04-leetcode101/WhiteNight.md

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>```java
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>/**
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> *给定一个二叉树，检查它是否是镜像对称的。
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> *
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> */
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>public class Tree9
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>{
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> public static class TreeNode
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> {
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> int data;
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> TreeNode left;
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> TreeNode right;
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>
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> TreeNode(int val) {
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> data = val;
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> }
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> }
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>
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> public boolean isSymmetric(TreeNode root)
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> {
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> if (root == null)
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> return true;
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> return checkIsSymmetric(root.left, root.right);
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> }
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>
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> public static boolean checkIsSymmetric(TreeNode leftNode, TreeNode rightNode)
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> {
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> if (leftNode == null && rightNode == null)
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> return true;
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> if ((leftNode == null && rightNode != null) || (leftNode != null && rightNode == null))
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> return false;
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> if (leftNode.data != rightNode.data)
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> return false;
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> return checkIsSymmetric(leftNode.left, rightNode.right) && checkIsSymmetric(leftNode.right, rightNode.left);
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> }
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>
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> public static void main(String[] args)
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> {
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> TreeNode p = new TreeNode(1);
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> p.left = new TreeNode(2);
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> p.right = new TreeNode(2);
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> p.left.left = new TreeNode(3);
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> p.right.right = new TreeNode(3);
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>
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> TreeNode q = new TreeNode(1);
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> q.left = new TreeNode(2);
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> q.right = new TreeNode(2);
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> q.left.right = new TreeNode(3);
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> q.right.right = new TreeNode(3);
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>
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> Tree9 t = new Tree9();
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> System.out.println(t.isSymmetric(p));
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> System.out.println(t.isSymmetric(q));
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>
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> }
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>
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>}
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>```

‎2018.12.04-leetcode101/fish.md

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```
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class Solution {
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public boolean isSymmetric(TreeNode treeNode) {
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if (treeNode == null) {
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return true;
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}else {
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return isSymmetric(treeNode.getLeft(), treeNode.getRight());
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}
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}
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public boolean isSymmetric(TreeNode left, TreeNode right) {
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if (left == null && right == null) {
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return true;
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}
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if (left == null || right == null) {
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return false;
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}
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if (left.getVal() != right.getVal()) {
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return false;
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}
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return isSymmetric(left.getLeft(), right.getRight()) && isSymmetric(left.getRight(), right.getLeft());
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}
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}
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class TreeNode{
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private int val;
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private TreeNode left;
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private TreeNode right;
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public TreeNode(int val, TreeNode left, TreeNode right) {
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this.val = val;
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this.left = left;
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this.right = right;
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}
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public int getVal() {
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return val;
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}
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public void setVal(int val) {
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this.val = val;
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}
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public TreeNode getLeft() {
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return left;
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}
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public void setLeft(TreeNode left) {
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this.left = left;
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}
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public TreeNode getRight() {
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return right;
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}
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public void setRight(TreeNode right) {
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this.right = right;
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}
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}

‎2018.12.04-leetcode101/kiritocly.md

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private static boolean solution(TreeNode root) {
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if (root == null) {
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return true;
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} else {
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//二叉树对称的条件
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//（1）左子树和右子树对称
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//（2）左子树的左子树和右子树的右子树对称
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//（3）左子树的右子树和右子树的左子树对称
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//递归判断
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return judge(root.leftNode, root.rightNode);
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}
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}
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private static boolean judge(TreeNode leftNode, TreeNode rightNode) {
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if (leftNode == null && rightNode == null) {
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return true;
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} else if (leftNode == null || rightNode == null) {
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return false;
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}
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//条件（1）
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if (leftNode.value != rightNode.value) {
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return false;
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}
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//条件（2）
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boolean flag1 = judge(leftNode.leftNode, rightNode.rightNode);
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//条件（3）
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boolean flag2 = judge(leftNode.rightNode, rightNode.leftNode);
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if (flag1 && flag2) {
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return true;
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}
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return false;
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}

‎2018.12.04-leetcode101/syuan.md

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##### 题目
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```
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给定一个二叉树，检查它是否是镜像对称的。
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例如，二叉树 [1,2,2,3,4,4,3] 是对称的。
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1
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/ \
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2 2
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/ \ / \
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3 4 4 3
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但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:
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1
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/ \
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2 2
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\ \
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3 3
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```
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##### 代码
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递归
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```
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public boolean isSymmetric(TreeNode root) {
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if (root==null) {
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return true;
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}else{
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return isSymmetric(root.left,root.right);
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}
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}
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public boolean isSymmetricCore(TreeNode leftNode, TreeNode rightNode){
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if (leftNode==null && rightNode==null) {
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return true;
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}
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if (leftNode.val!=rightNode.val || leftNode==null || rightNode==null) {
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return false;
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}
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return isSymmetricCore(leftNode.left,rightNode.right) && isSymmetricCore(leftNode.right,rightNode.left);
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}
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```
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##### 非递归法
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```
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class Solution {
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public boolean isSymmetric(TreeNode root) {
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if(root == null) return true;
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Stack<TreeNode> stack = new Stack<TreeNode>();
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stack.push(root.left);
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stack.push(root.right);
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while(!stack.empty()) {
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TreeNode right = stack.pop();
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TreeNode left = stack.pop();
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if (left == null && right == null) continue;
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else if (left == null || right == null || right.val != left.val) return false;
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stack.push(left.left);
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stack.push(right.right);
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stack.push(left.right);
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stack.push(right.left);
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}
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return true;
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}
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}
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```

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