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; }