题解java| #重建二叉树#

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
通过分割数组  递归实现，前序 1 2473568  中序 472 1 5386 这样就获得第一个节点1，然后1的左子树前序 247与中序  472再递归成 2 47 ，47 2 一直分隔到只有一个节点
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre,int [] vin) {
         return fenge(pre,vin,0,pre.length-1,0, vin.length-1);
    }
     static public TreeNode fenge(int [] pre,int [] vin,int left,int right,int left1,int right1) {
        if(pre.length==0){
            return null;
        }
        TreeNode node=null;
        if (left<=right &&left1<=right1 &&left<pre.length&&left1<vin.length) {
            node = new TreeNode(pre[left]);
            int count = 0;
            while (vin[count+left1] != node.val) {
                count++;
//
            }
            TreeNode leftNode = fenge(pre, vin, left + 1, count + left, left1, count + left1 - 1);
            TreeNode rightNode = fenge(pre, vin, count + left + 1, right, count + left1 + 1, right1);
            node.left=leftNode;
            node.right=rightNode;

        }
        return  node ;
//        new TreeNode(pre[left])
    }
}