Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Example 1:
Input: [1,3,null,null,2] 1 / 3 \ 2 Output: [3,1,null,null,2] 3 / 1 \ 2Example 2:
Input: [3,1,4,null,null,2] 3 / \ 1 4 / 2 Output: [2,1,4,null,null,3] 2 / \ 1 4 / 3
Follow up:
- A solution using O(n) space is pretty straight forward.
- Could you devise a constant space solution?
恢复二叉搜索树。
题目即是题意。给的条件是一个不完美的BST,其中有两个节点的位置摆放反了,请将这个BST恢复。
这个题的思路其实很简单,就是中序遍历。followup不需要额外空间的做法我暂时不会做,日后再补充。既然是BST,中序遍历最管用了。中序遍历BST,正常情况是返回一个升序的数组,所以遍历过程中的相邻节点应该也是递增的。如果找到某个节点B小于等于他之前遍历到的那个节点A,则证明AB是那两个有问题的节点。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public void recoverTree(TreeNode root) { 3 TreeNode pre = null; 4 TreeNode cur = root; 5 TreeNode first = null; 6 TreeNode second = null; 7 Stack<TreeNode> stack = new Stack<>(); 8 while (!stack.isEmpty() || cur != null) { 9 while (cur != null) { 10 stack.push(cur); 11 cur = cur.left; 12 } 13 cur = stack.pop(); 14 if (pre != null && cur.val <= pre.val) { 15 if (first == null) { 16 first = pre; 17 } 18 second = cur; 19 } 20 pre = cur; 21 cur = cur.right; 22 } 23 int temp = first.val; 24 first.val = second.val; 25 second.val = temp; 26 } 27 }