题解 | 重建二叉树

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) {
        if(preOrder.length==0){
          return null;
        }
       int rootValue= preOrder[0];
       TreeNode  root= new TreeNode(rootValue);
       for(int i=0;i<vinOrder.length;i++){
           if(vinOrder[i]==rootValue){
            int[] left1= Arrays.copyOfRange(vinOrder,0,i);
            int[] right1= Arrays.copyOfRange(vinOrder,i+1,preOrder.length);
            int[] left2=  Arrays.copyOfRange(preOrder,1,i+1);
            int[] right2=  Arrays.copyOfRange(preOrder,i+1,preOrder.length);
            root.left=reConstructBinaryTree(left2,left1);
            root.right=reConstructBinaryTree(right2,right1);
            break;
           }
       }
        return root;
        // write code here
    }
}

每次递归传入分割完好的左子树的前序遍历数组和中序遍历数组。