BST to double linkedlist in inorder traversal sequence
Follow Up: 如果这个BST自带prev, next, 给一个value,插进去这个node,并补全left,right,prev,next
1 class TreeNode { 2 public int val; 3 TreeNode left; 4 TreeNode right; 5 } 6 7 TreeNode newRoot 8 9 TreeNode tree2Dll(TreeNode root) { 10 TreeNode pre = null; 11 TreeNode newRoot = null; 12 helper(root, pre); 13 return newRoot; 14 } 15 16 public void helper(TreeNode cur, TreeNode pre) { 17 if (cur == null) { 18 return; 19 } 20 helper(cur.left, pre); 21 cur.left = pre; 22 if (pre == null) { 23 newRoot = cur; 24 } 25 else pre.right = cur; 26 pre = cur; 27 helper(cur.right, pre, newRoot); 28 }
Follow up:
第一步完成基础上,把这个new value插入BST中,一边插入一边记录沿途的大于value的root node(即走了该root node的left child),作为inorder successor. 找到inorder successor之后,通过其prev指针找到inorder predecessor, 改指针就好了