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[LeetCode] 897. Increasing Order Search Tree

Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

Example 1:

Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

Example 2:

Input: root = [5,1,7]
Output: [1,null,5,null,7]

Constraints:

The number of nodes in the given tree will be in the range [1, 100].
0 <= Node.val <= 1000

递增顺序搜索树。

给你一个树，请你按中序遍历重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。

这道题跟之前的114题很像，也是属于需要把树做扁平化处理的题目。既然题目都说了用中序遍历，我这里给出两种做法，迭代和递归。时间空间复杂度均是 O(n)。代码应该是 self-explained，只是需要注意当处理节点的时候，需要找到新的 head 节点，以及把处理过的每个节点的左指针设置成 NULL，新的树里面每个节点都只有右指针。

迭代Java实现

 1 class Solution {
 2     public TreeNode increasingBST(TreeNode root) {
 3         // corner case
 4         if (root == null) {
 5             return null;
 6         }
 7         
 8         // normal case
 9         Deque<TreeNode> stack = new ArrayDeque<>();
10         TreeNode cur = root;
11         TreeNode pre = null;
12         TreeNode head = null;
13         while (!stack.isEmpty() || cur != null) {
14             while (cur != null) {
15                 stack.push(cur);
16                 cur = cur.left;
17             }
18             cur = stack.pop();
19             if (head == null) {
20                 head = cur;
21             }
22             cur.left = null;
23             if (pre != null) {
24                 pre.right = cur;
25             }
26             pre = cur;
27             cur = cur.right;
28         }
29         return head;
30     }
31 }

递归Java实现

 1 class Solution {
 2     TreeNode pre = null;
 3     TreeNode head = null;
 4 
 5     public TreeNode increasingBST(TreeNode root) {
 6         if (root == null) {
 7             return null;
 8         }
 9         increasingBST(root.left);
10         if (head == null) {
11             head = root;
12         }
13         if (pre != null) {
14             root.left = null;
15             pre.right = root;
16         }
17         pre = root;
18         increasingBST(root.right);
19         return head;
20     }
21 }