leetcode 105.从前序与中序遍历序列构造二叉树 Java

题目链接

https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

描述

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

示例

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

初始代码模板

/**
 * 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) {

    }
}

代码

推荐题解：https://labuladong.gitbook.io/algo/shu-ju-jie-gou-xi-lie/er-cha-shu-xi-lie-2
写的已经很好，不再赘述

/**
 * 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) {
        return build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
    }

    private TreeNode build(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
        if (preStart > preEnd) {
            return null;
        }

        int rootVal = preorder[preStart];
        int index = -1;

        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == rootVal) {
                index = i;
                break;
            }
        }

        int leftSize = index - inStart;

        TreeNode root = new TreeNode(rootVal);

        root.left = build(preorder, preStart + 1, preStart + leftSize, inorder, inStart, index - 1);
        root.right = build(preorder, preStart + leftSize + 1, preEnd, inorder, index + 1, inEnd);

        return root;
    }
}