题干：
根据一棵树的前序遍历与中序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
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解答：
/*** 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){return null ;}if(preorder.length == 1){return new TreeNode(preorder[0]);}TreeNode root = new TreeNode(preorder[0]);int rootNum = 0 ;for(int i =0 ; i < inorder.length; i ++){if(inorder[i] == preorder[0]){rootNum = i;break;}}int[] inLeft = Arrays.copyOfRange(inorder,0,rootNum);int[] preLeft = Arrays.copyOfRange(preorder,1,1+inLeft.length);int[] inRight = Arrays.copyOfRange(inorder,rootNum+1,inorder.length);int[] preRight = Arrays.copyOfRange(preorder,inorder.length-inRight.length,inorder.length);root.left = buildTree(preLeft,inLeft);root.right = buildTree(preRight,inRight);return root;}
}
