[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;
     }