使用排序数组/链表/preorder构建二叉搜索树

2018-08-13 11:29:05

一、Convert Sorted Array to Binary Search Tree

问题描述：

问题求解：

    public TreeNode sortedArrayToBST2(int[] nums) {
        if (nums == null || nums.length == 0) {
            return null;
        }
        return helper(0, nums.length - 1, nums);
    }

    private TreeNode helper (int start, int end, int[] nums) {
        if (start > end) {
            return null;
        }
        int mid= start + (end - start) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = helper(start, mid - 1, nums);
        root.right = helper(mid + 1, end , nums);
        return root;
    }

二、Convert Sorted List to Binary Search Tree

问题描述：

问题求解：

    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) return null;
        return helper(head, null);
    }

    private TreeNode helper(ListNode head, ListNode tail) {
        if (head == tail) return null;
        ListNode mid = getMid(head, tail);
        TreeNode root = new TreeNode(mid.val);
        root.left = helper(head, mid);
        root.right = helper(mid.next, tail);
        return root;
    }

    private ListNode getMid(ListNode head, ListNode tail) {
        ListNode slow = head;
        ListNode fast = head;
        while (fast != tail && fast.next != tail) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

三、Construct Binary Search Tree from Preorder Traversal

问题描述：

问题求解：

    public TreeNode bstFromPreorder(int[] preorder) {
        if (preorder == null || preorder.length == 0) return null;
        return helper(preorder, 0, preorder.length - 1);
    }

    private TreeNode helper(int[] preorder, int left, int right) {
        if (right < left) return null;
        if (left == right) {
            TreeNode node = new TreeNode(preorder[left]);
            node.left = null;
            node.right = null;
            return node;
        }
        TreeNode node = new TreeNode(preorder[left]);
        int mid = getMid(preorder, left, right);
        node.left = helper(preorder, left + 1, mid - 1);
        node.right = helper(preorder, mid, right);
        return node;
    }

    private int getMid(int[] nums, int left, int right) {
        int res = left + 1;
        for (; res <= right; res++) {
            if (nums[res] > nums[left]) break;
        }
        return res;
    }