Structure binary tree from preorder and inorder traversal for leetcode topic analysis

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

Note:

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

Subject requirements ： According to the preorder traversal and inorder traversal sequence , Building a binary tree .

• The former sequence traversal p：1 2 4 3 5 6
• In the sequence traversal i：4 2 1 5 3 6

    1
   / \
  2   3
 /   / \
4   5   6

The former sequence traversal p0 Is the root node of a binary tree ,1 yes i2, be ：

• p1...2 yes 1 Preorder traversal of the left subtree of ,i0...1 yes 1 The middle order traversal of the left subtree of
• p3...5 yes 1 Preorder traversal of right subtree of ,i3...5 yes 1 The middle order traversal of the right subtree of
• recursive

int p = 0;
    int[] preorder;
    int[] inorder;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        this.preorder = preorder;
        this.inorder = inorder;
        return buildTree(0, preorder.length);
    }
    TreeNode buildTree(int start, int end) {
        if (start >= end) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[p]);
        int i;
        for (i = start; i < end && preorder[p] != inorder[i]; i++)
            ;
        p++;
        root.left = buildTree(start, i);
        root.right = buildTree(i + 1, end);
        return root;
    }