重建二叉树——剑指offer

题目描述

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。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。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}，则重建二叉树并返回。

这种解法要复制数组，效率低！！！看后面别人提供的更好思路

public class Main8 {
public static TreeNode reConstructBinaryTree(int [] pre,int [] in) {
if(pre == null || pre.length ==0|| in==null || in.length==0)
return null;
TreeNode root=new TreeNode(pre[0]);
int i=0;
for(;i<in.length;i++){
if(in[i]==pre[0])
break;
}
int[] pre1=new int[i];
int[] pre2=new int[in.length-i-1];
System.arraycopy(pre, 1, pre1,0, i);
System.arraycopy(pre, i+1, pre2,0, in.length-i-1);

int[] in1=new int[i];
int[] in2=new int[in.length-i-1];
System.arraycopy(in, 0, in1, 0, i);
System.arraycopy(in, i+1, in2, 0, in.length-i-1);

root.left=reConstructBinaryTree(pre1, in1);
root.right=reConstructBinaryTree(pre2, in2);
return root;
}
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};
TreeNode tree=reConstructBinaryTree(pre,in);
}

}

这种方法不用复制数组：

public class Main8 {
public static TreeNode reConstructBinaryTree(int [] pre,int [] in) {
TreeNode root=reConstructBinaryTree(pre,0,pre.length-1,in,0,in.length-1);
return root;
}
private static TreeNode reConstructBinaryTree(int[] pre,int startPre,int endPre,int[] in,int startIn,int endIn)
{
if(pre==null || in==null)
return null;
if(startIn>endIn || startPre>endPre)
return null;
TreeNode root=new TreeNode(pre[startPre]);
for(int i=startIn;i<=endIn;i++){
if(in[i]==pre[startPre]){
root.left=reConstructBinaryTree(pre,startPre+1,startPre+i-startIn,in,startIn,i-1);
root.right=reConstructBinaryTree(pre,startPre+i-startIn+1,endPre,in,i+1,endIn);
}
}
return root;
}
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};
TreeNode tree=reConstructBinaryTree(pre,in);
}

}