105. 从前序与中序遍历序列构造二叉树(二叉树，中等)

题目链接

给定一棵树的前序遍历 preorder 与中序遍历 inorder。请构造二叉树并返回其根节点。

示例 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

示例 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

提示:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder 和 inorder 均无重复元素
• inorder 均出现在 preorder
• preorder 保证为二叉树的前序遍历序列
• inorder 保证为二叉树的中序遍历序列

解答

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return buildTreeWithPointers(preorder, inorder, 0, preorder.length, 0, inorder.length);
    }

    // 用指针记录用于建树的先序与中序的数组段，以建树
    public TreeNode buildTreeWithPointers(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd) {
        if (preorder == null || preStart >= preorder.length || preStart >= preEnd) return null;
        if (preStart+1 == preEnd) return new TreeNode(preorder[preStart]);
        TreeNode root = new TreeNode(preorder[preStart]);
        for (int i = 0; i < preEnd-preStart; i++) {
            if (inorder[inStart+i] == root.val) {
                root.left = buildTreeWithPointers(preorder, inorder, preStart+1, preStart+i+1, inStart, inStart+i);
                root.right = buildTreeWithPointers(preorder, inorder, preStart+i+1, preEnd, inStart+i+1, inEnd);
                break;
            }
        }
        return  root;
    }
}