输入某二叉树的前序遍历和中序遍历的结果，请重建该二叉树。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，给出
前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：
3
/ \
9 20
/ \
15 7
限制：
0 <= 节点个数 <= 5000
代码：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if(preorder.length==0||inorder.length==0){
return null;
}
Map<Integer,Integer> map = new HashMap<>();
for(int i=0;i<inorder.length;i++){
map.put(inorder[i],i);
}
TreeNode node = find(preorder,0,preorder.length-1,inorder,0,inorder.length-1,map);
return node;
}
public TreeNode find(int preorder[],int prestart,int preend,int inorder[],int instart,int inend,Map<Integer,Integer> map)
{
if(prestart>preend){
return null;
}
TreeNode node = new TreeNode(preorder[prestart]);
if(preend==prestart){
return node;
}else{
int rootindex = map.get(preorder[prestart]);
int leftnodes = rootindex-instart;//左子树多少个节点
int rightnodes = inend-rootindex;//右子树多少个节点
TreeNode left = find(preorder,prestart+1,prestart+leftnodes,inorder,instart,rootindex-1,map);
TreeNode right = find(preorder,preend-rightnodes+1,preend,inorder,rootindex+1,inend,map);
node.left = left;
node.right = right;
return node;
}
}
}
