leetcode-99.-恢复二叉搜索树

描述

二叉搜索树中的两个节点被错误地交换。

请在不改变其结构的情况下，恢复这棵树。

示例 1:

输入: [1,3,null,null,2]

   1
  /
 3
  \
   2

输出: [3,1,null,null,2]

   3
  /
 1
  \
   2
示例 2:

输入: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

输出: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3
进阶:

　

解法：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public void recoverTree(TreeNode root) {
        List<Integer> nums = new ArrayList<>();
        inOrder(root, nums);
        int[] swaped = findTwoSwap(nums);
        recover(root, 2, swaped[0], swaped[1]);
    }

    private void inOrder(TreeNode root, List<Integer> nums) {
        if (root == null) {
            return;
        }
        inOrder(root.left, nums);
        nums.add(root.val);
        inOrder(root.right, nums);
    }

    private int[] findTwoSwap(List<Integer> nums) {
        int x = -1;
        int y = -1;
        for (int i = 0; i < nums.size() - 1; i++) {
            if (nums.get(i+1) < nums.get(i)) {
                y = nums.get(i+1) ;
                if (x == -1) {
                    x = nums.get(i);
                } else {
                    break;
                }
            }
        }

        return new int[] {x, y};
    }

    public void recover(TreeNode r, int count, int x, int y) {

        if (count == 0) {
            return;
        }

        if (r == null) {
            return;
        }

     if (r.val == x || r.val == y) {
        if (r.val == x) {
            r.val = y;
        } else if (r.val != x) {
            r.val = x;
        }
        count--;
      }
      recover(r.left, count, x, y);
      recover(r.right, count, x, y);
  }    
}

　　

时间复杂度： N， 空间复杂度： N

 

参考： https://leetcode-cn.com/problems/validate-binary-search-tree/