题解 | #重建二叉树#

import java.util.*;
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre, int [] vin) {
        if (pre == null || pre.length == 0 || vin == null || vin.length == 0) {
            return null;
        }
        // 构造二叉树
        TreeNode root = constructBinaryTree(pre, 0, pre.length - 1, vin, 0,
                                            vin.length - 1);
        return root;

    }

    /**
    * 传入前序序列数组，起点和终点
    * 传入中序序列数组，起点和终点
    * */
    private TreeNode constructBinaryTree(int[] pre, int startPre, int endPre,
                                         int[] in, int startIn, int endIn) {
        // 递归终止条件
        if (startPre > endPre || startIn > endIn) {
            return null;
        }

        // 构建根节点，前序第一个值
        TreeNode root = new TreeNode(pre[startPre]);
        // 遍历中序，找到根节点
        for (int index = startIn; index <= endIn; index++) {
            if (in[index] == pre[startPre]) {
                // 递归构建左子树，左边数组列表的起终点结算
                root.left = constructBinaryTree(pre, startPre + 1, startPre + (index - startIn),
                                                in, startIn, index - 1);
                // 递归构建右子树，右边数组列表的起终点结算
                root.right = constructBinaryTree(pre, (index - startIn) + startPre + 1, endPre,
                                                 in, index + 1, endIn);
                break;
            }
        }
        return root;
    }
}

解决思想：递归子树遍历，主要注意每段子树的边界索引值