从前序与中序遍历序列构造二叉树

从前序与中序遍历序列构造二叉树

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]
示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-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

解题思路：
本题与从中序与后序遍历构造二叉树的思路是一致的，具体分析建议看我的上一篇文章从中序与后序遍历序列构造二叉树，本题直接贴出代码：

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return build(preorder, inorder);
    }
    public TreeNode build(int[] preorder, int[] inorder) {
        if (preorder == null || inorder == null) {return null;}
        if (preorder.length == 0 || inorder.length == 0) {return null;}
        if (preorder.length == 1) {return new TreeNode(preorder[0]);}
        TreeNode root = new TreeNode(preorder[0]);
        int index = 0;
        for (; index < inorder.length; index++) {
            if (inorder[index] == preorder[0]) {
                break;
            }
        }
        root.left = build(Arrays.copyOfRange(preorder, 1, index + 1), Arrays.copyOfRange(inorder, 0, index));
        root.right = build(Arrays.copyOfRange(preorder, index + 1, preorder.length), Arrays.copyOfRange(inorder, index + 1, inorder.length));
        return root;
    }
}

class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return build(preorder, 0, preorder.length, inorder, 0, inorder.length);
    }
    public TreeNode build(int[] preorder, int pLeft, int pRight, int[] inorder, int iLeft, int iRight) {
        //没有元素
        if (pRight - pLeft < 1) {return null;}
        //只有一个元素
        if (pRight - pLeft == 1) {return new TreeNode(preorder[pLeft]);}
        TreeNode root = new TreeNode(preorder[pLeft]);
        int index = iLeft;
        for (; index < iRight; index++) {
            if (inorder[index] == preorder[pLeft]) {
                break;
            }
        }
        root.left = build(preorder, pLeft + 1, index + pLeft - iLeft + 1, inorder, iLeft, index);
        root.right = build(preorder, index + pLeft - iLeft + 1, pRight, inorder, index + 1, iRight);
        return root;
    }
}