LeetCode：重建二叉树

一、题目描述

二、思路分析

2.1 分析

首先前序遍历的遍历顺序是根节点 -> 左节点 -> 右节点，而中序遍历的遍历顺序是左节点 -> 根节点 -> 右节点。我们可以从二叉树的前序遍历确定根节点的位置，进而在中序遍历中根据根节点划分左右子树范围。

2.2 图解

• 参考 BonFire 的题解

三、题解

/**
 * 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) {
        if (preorder == null || preorder.length == 0) {
            return null;
        }
        List<Integer> preorderList = new ArrayList<>();
        List<Integer> inorderList = new ArrayList<>();
        for (int i = 0; i < preorder.length; i++) {
            preorderList.add(preorder[i]);
            inorderList.add(inorder[i]);
        }

        return builder(preorderList, inorderList);
    }

    private TreeNode builder(List<Integer> preorderList, List<Integer> inorderList) {
        if (inorderList.isEmpty()) {
            return null;
        }
        // 取出根节点的值
        int rootVal = preorderList.remove(0);
        // 确定根节点的下标位置
        int rootIndex = inorderList.indexOf(rootVal);
        // 构建二叉树
        TreeNode root = new TreeNode(rootVal);
        root.left = builder(preorderList, inorderList.subList(0, rootIndex));
        root.right = builder(preorderList, inorderList.subList(rootIndex + 1, inorderList.size()));
        return root;
    }
}