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add 重构二叉树
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‎Algorithm/剑指offer/coding-interviews/.idea/workspace.xml‎

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‎Algorithm/剑指offer/coding-interviews/pom.xml‎

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<groupId>com.leosanqing</groupId>
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<artifactId>coding-interviews</artifactId>
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<version>1.0-SNAPSHOT</version>
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<build>
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<plugins>
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<plugin>
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<groupId>org.apache.maven.plugins</groupId>
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<artifactId>maven-compiler-plugin</artifactId>
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<configuration>
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<source>6</source>
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<target>6</target>
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</configuration>
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</plugin>
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</plugins>
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</build>
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</project>

‎Algorithm/剑指offer/coding-interviews/src/main/java/com/leosanqing/algorithm/_6_PrintListFromTailToHead.java‎

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* 使用递归来打印
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*
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* 如果像书上使用其他辅助容器，那基本所有容器都可以。
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* @param head
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* @param head 头结点
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*/
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private static void method(Node head) {
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if (head != null) {

‎Algorithm/剑指offer/coding-interviews/src/main/java/com/leosanqing/algorithm/_7_ReConstructBinaryTree.java‎

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package com.leosanqing.algorithm;
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import java.util.Arrays;
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/**
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* @Author: leosanqing
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* @Date: 2019-11-09 11:44
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*
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*
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* 输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。
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*
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* 假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
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* 例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，
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* 则重建二叉树并返回头结点。
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*
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*/
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public class _7_ReConstructBinaryTree {
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public static void main(String[] args) {
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int[] pre={1,2,4,7,3,5,6,8};
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int[] in ={4,7,2,1,5,3,8,6};
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System.out.println(method(pre,in).value);
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}
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/**
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* ！！！！！注意：题目中说，数组中的数字是不会重复的。不然用不了下面的方法
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*
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* 这道题主要运用 二叉树先序遍历和中序遍历的特点
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*
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* 1.先序遍历中，第一个就是根节点。按照上面的例子，也就是 1
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* 2.然后我们再去中序遍历的 数组中找到这个 1 。
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* 在1前面的那部分就是 这个根节点的左子树{4,7,2}。右边的为右子树{5,3,8,6}
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* 3.然后我们再在左子树中找到第一个数 4,再重复上述步骤
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*
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* 这样我们就能找到树的结构，然后一直递归。再在这个过程中添加上左右子节点。就能完成重构了
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*
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* @param pre 左子树
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* @param in 右子树
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* @return 根节点
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*/
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private static TreeNode method(int[] pre,int[] in){
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if(null == pre||null == in){
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return null;
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}
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if(pre.length <=0||in.length<=0) {
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return null;
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}
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if(pre.length!=in.length){
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return null;
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}
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TreeNode root = new TreeNode(pre[0]);
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for (int i = 0; i < pre.length; i++) {
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// 找到中序遍历的根节点
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if(pre[0] == in[i]){
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root.left = method(Arrays.copyOfRange(pre,i,i+1),Arrays.copyOfRange(in,0,i));
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root.right = method(Arrays.copyOfRange(pre,i+1,pre.length),Arrays.copyOfRange(in,i+1,in.length));
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// root.left = reConstructBinaryTree(
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// Arrays.copyOfRange(pre,1,i+1),Arrays.copyOfRange(in,0,i));
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// root.right = reConstructBinaryTree(
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// Arrays.copyOfRange(pre,i+1,pre.length),Arrays.copyOfRange(in,i+1,in.length));
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}
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}
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return root;
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}
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/**
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* 树结构
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*/
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static class TreeNode{
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int value;
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TreeNode left;
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TreeNode right;
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public TreeNode(int value) {
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this.value = value;
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}
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}
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}

‎Algorithm/剑指offer/coding-interviews/target/classes/com/leosanqing/algorithm/FindNumber.class‎

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‎Algorithm/剑指offer/coding-interviews/target/classes/com/leosanqing/algorithm/_2_Singleton$1.class‎

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‎Algorithm/剑指offer/coding-interviews/target/classes/com/leosanqing/algorithm/_2_Singleton$Hungry.class‎

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‎Algorithm/剑指offer/coding-interviews/target/classes/com/leosanqing/algorithm/_2_Singleton$Lazy.class‎

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‎Algorithm/剑指offer/coding-interviews/target/classes/com/leosanqing/algorithm/_2_Singleton$Singleton.class‎

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‎Algorithm/剑指offer/coding-interviews/target/classes/com/leosanqing/algorithm/_2_Singleton$StaticSingleton$StaticSingletonHolder.class‎

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