【LeetCode】99.Recover Binary Search Tree(Hard)解题报告

【LeetCode】99.Recover Binary Search Tree(Hard)解题报告

题目地址：https://leetcode.com/problems/recover-binary-search-tree/description/
题目描述：

  Two elements of a binary search tree (BST) are swapped by mistake.
  Recover the tree without changing its structure.
  Note:A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.

  Your algorithm should run in O(n) time and uses constant space.
  通过这道题，要学习marris traversal这种方法。

Solution:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
/*
1 5 3 4 2 6
boolean flag
preNode.val > curNode.node
first = pre
second = cur

Marris traversal:建桥和拆桥
中序遍历应该是由小到大

*/
class Solution {
    public void recoverTree(TreeNode root) {
        TreeNode first = null;
        TreeNode second = null;
        boolean firstTime = true;
        TreeNode pre = new TreeNode(Integer.MIN_VALUE);
        //中序的marris traversal
        while(root != null){
            if(root.left != null){
                TreeNode temp = root.left;
                //目的是走到左子树最右边的node
                while(temp.right != null && temp.right != root){
                    temp = temp.right;
                }
                //第一次到达左子树最右边的那个node
                if(temp.right == null){
                    //建桥
                    temp.right = root;
                    //把root向左边移动一个
                    root = root.left;
                }else{ //已经建过桥了，拆桥
                    temp.right = null;
                    //visit root.val
                    if(pre.val > root.val && firstTime){
                        first = pre;
                        firstTime = false;
                    }
                    if(pre.val > root.val && !firstTime){
                        second = root;
                    }
                    pre = root;
                    root = root.right;
                }
            }else{ //到了最左边的node，visit root.val
                if(pre.val > root.val && firstTime){
                    first = pre;
                    firstTime = false;
                }
                if(pre.val > root.val && !firstTime){
                    second = root;
                }
                pre = root;
                root = root.right;
            }
        }

        if(first!=null && second!=null){
            int tmp = first.val;
            first.val = second.val;
            second.val = tmp;
        }

    }
}

Date：2017年11月23日