剑指offer 面试题6 重建二叉树 java版答案

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package offerAnswer;/* 定义二叉树*/public class BinaryTreeNode {    int value;    BinaryTreeNode left;    BinaryTreeNode right;}

package OfferAnswer;/** * 面试题6 * 重建二叉树 * @author lwk * 1.通过前序遍历确定根节点 * 2.在中序遍历找到该节点，那么在该节点前面的为左子树的节点，该节点后面的为右子树的节点 * 3.递归 */public class Answer06 {     public static void main(String[] args) {     int[] pre={1,2,4,7,3,5,6,8};  int[] in = {4,7,2,1,5,3,8,6};  BinaryTreeNode root = reconstructBinaryTree(pre,in);  BinaryTreeNode p = root;         while(null != p){         System.out.println(p.value);         p = p.left;         } }          public static BinaryTreeNode reconstructBinaryTree(int[] pre, int[] in){     if(pre == null || pre.length == 0 || in == null || in.length == 0){     return null;     }     return reconstructBinaryTree(pre,0,in,0,pre.length-1);     }          public static BinaryTreeNode reconstructBinaryTree(int[] pre, int preStart, int[] in, int inStart, int inEnd){     if(inStart > inEnd){     return null;     }     BinaryTreeNode root = new BinaryTreeNode();     root.value = pre[preStart];     if(inStart == inEnd){     return root;     }     //在中序遍历中寻找根节点     int i = inStart;     for (i = inStart; i < in.length; i++) { if(in[i] == root.value){ break; } }     //下标i即为根节点的位置     //一分为二，寻找子数组的根节点，分别为该根节点的左右子节点     root.left = reconstructBinaryTree(pre, preStart+1 , in, inStart, i-1);     root.right = reconstructBinaryTree(pre, i-inStart+preStart+1, in, i+1, inEnd);     return root;     }     }

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