[Leetcode][JAVA] Convert Sorted Array to Binary Search Tree && Convert Sorted List to Binary Search Tree

Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

很简单的二分法，只要给出Array的开始和结束下标作为参数传入即可。

 public TreeNode sortedArrayToBST(int[] num) {
        return constructBST(num,0,num.length-1);
    }
    public TreeNode constructBST(int[] num, int start, int end) {
        if(start>end)
            return null;
        if(start==end)
            return new TreeNode(num[start]);
        int mid = (start+end)/2;
        TreeNode root = new TreeNode(num[mid]);
        root.left = constructBST(num, start, mid-1);
        root.right = constructBST(num, mid+1, end);
        return root;
    }

Convert Sorted List to Binary Search Tree

 Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

对于链表来说，情况稍微复杂一些，最简单的思路是可以遍历一遍链表存到array中然后构建。这里采用直接构建的方法，就是用快慢指针找到链表中点，左半部分链表终点指向null，递归构建成左子树。右半部分递归构建成右子树即可。如果不想破坏原有链表那么可以采用先存到array中再构建的方法。

    public TreeNode sortedListToBST(ListNode head) {
        if(head==null)
            return null;
        if(head.next == null)
            return new TreeNode(head.val);
        ListNode fast = head.next;
        ListNode slow = head;
        while(fast!=null && fast.next!=null && fast.next.next!=null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        ListNode root = slow.next;
        slow.next = null;
        TreeNode tRoot = new TreeNode(root.val);
        tRoot.left = sortedListToBST(head);
        tRoot.right = sortedListToBST(root.next);
        return tRoot;
    }