题解 | #重建二叉树#
重建二叉树
https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param preOrder int整型一维数组 * @param vinOrder int整型一维数组 * @return TreeNode类 */ public TreeNode reConstructBinaryTree (int[] preOrder, int[] vinOrder) { // write code here if (preOrder.length == 0 || vinOrder.length == 0) { return null; } return buildTree(preOrder, 0, preOrder.length - 1, vinOrder, 0, vinOrder.length - 1); } public TreeNode buildTree(int[] preOrder, int prestart, int preend, int[] vinOrder, int vinstart, int vinend) { if (prestart > preend || vinstart > vinend) { return null; } TreeNode head = new TreeNode(preOrder[prestart]); int rootindex = vinstart; while(rootindex <= vinend && vinOrder[rootindex] != preOrder[prestart]) { rootindex++; } head.left = buildTree(preOrder, prestart + 1, prestart + rootindex - vinstart, vinOrder, vinstart, rootindex - 1); head.right = buildTree(preOrder, prestart + rootindex - vinstart + 1, preend, vinOrder, rootindex + 1, vinend); return head; } }