根據一棵樹的前序遍歷與中序遍歷構造二叉樹。注意:
你可以假設樹中沒有重復的元素。例如，給出前序遍歷 preorder = [3,9,20,15,7]
中序遍歷 inorder = [9,3,15,20,7]
返回如下的二叉樹：3/ \9  20/  \15   7

代碼

/*** 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) {return build(preorder,0,preorder.length-1,inorder,0,inorder.length-1);}public TreeNode build(int[] preorder,int pl,int pr, int[] inorder,int il,int ir) {if(pl>pr||il>ir) return null;//遞歸邊界int index=0;while (preorder[pl]!=inorder[il+index]) index++;//找出父節點在中序遍歷的位置TreeNode temp=new TreeNode(preorder[pl]);temp.left=build(preorder, pl+1, pl+index, inorder, il, il+index-1);
//根據index和父節點將兩個數組分成兩半，遞歸尋找左右子樹temp.right=build(preorder, pl+index+1, pr, inorder, il+index+1, ir);return temp;}
}