LeetCode 105 Construct Binary Tree from Preorder and Inorder Traversal (DFS 构造)

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Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

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

题目分析：根据先序找根，根据中序二分左右子树

5ms，时间击败64%

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    public TreeNode buildHelper(int[] preorder, int pos, int[] inorder, int left, int right) {
        TreeNode root = new TreeNode(preorder[pos]);
        if (left == right) {
            return root;
        }
        for (int i = 0; i < inorder.length; i++) {
            if (preorder[pos] == inorder[i]) {
                int leftSize = i - left;
                if (pos + 1 < inorder.length && left <= i - 1) {
                    root.left = buildHelper(preorder, pos + 1, inorder, left, i - 1);
                }
                if (pos + leftSize + 1 < inorder.length && i + 1 <= right) {
                    root.right = buildHelper(preorder, pos + leftSize + 1, inorder, i + 1, right);
                }
                break;
            }
        }
        return root;
    }
    
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder.length == 0) {
            return null;
        }
        return buildHelper(preorder, 0, inorder, 0, inorder.length - 1);   
    }
}