LeetCode-106从中序与后序遍历序列构造二叉树-中等

标题：106从中序与后序遍历序列构造二叉树-中等

题目

示例1

输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]

示例2

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

提示

• 1 <= inorder.length <= 3000
• postorder.length == inorder.length
• -3000 <= inorder[i], postorder[i] <= 3000
• inorder 和 postorder 都由 不同 的值组成
• postorder 中每一个值都在 inorder 中
• inorder 保证是树的中序遍历
• postorder 保证是树的后序遍历

代码Java

// 时效较差，因为使用了很多Arrays方法
public TreeNode buildTree(int[] inorder, int[] postorder) {
    // 1. 后序空的话 直接设为null
    if (postorder.length == 0) return null;
    // 2. 选取后序的最后一个元素 作为本结点的值
    int endval = postorder[postorder.length - 1];
    TreeNode root = new TreeNode(endval);
    // 3. 找到inorder中的切割点
    int mid = -1;
    for (int i = 0; i < inorder.length; i++) {
        if (inorder[i] == endval) {
            mid = i;
            break;
        }
    }
    // 4. 切割中序
    int[] inLeft = Arrays.copyOfRange(inorder, 0, mid);
    int[] inRight = Arrays.copyOfRange(inorder, mid + 1, inorder.length);
    // 5. 寻找后序
    int[] postLeft = Arrays.copyOfRange(postorder, 0, inLeft.length);
    int[] postRight = Arrays.copyOfRange(postorder, postLeft.length, postorder.length - 1);
    root.left = buildTree(inLeft, postLeft);
    root.right = buildTree(inRight, postRight);
    return root;
}


public TreeNode buildTree1(int[] inorder, int[] postorder) {
    return build(inorder, 0, inorder.length, postorder, 0, postorder.length);
}
// 用双指针代替 Arrays方法
public TreeNode build(int[] inorder, int inLeft, int inRight, int[] postorder, int postLeft, int postRight) {
    if (postRight - postLeft < 1) {
        return null;
    }
    // 只有一个元素了
    if (inRight - inLeft == 1) {
        return new TreeNode(inorder[inLeft]);
    }
    int endval = postorder[postRight - 1];
    TreeNode root = new TreeNode(endval);
    int mid = -1;
    for (int i = inLeft; i < inRight; i++) {
        if (inorder[i] == endval) {
            mid = i;
            break;
        }
    }
    root.left = build(inorder, inLeft, mid,
            postorder, postLeft, postLeft + mid - inLeft);
    root.right = build(inorder, mid + 1, inRight,
            postorder, postLeft + mid - inLeft, postRight - 1);
    return root;
}