【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 6boolean flagpreNode.val > curNode.nodefirst = presecond = curMarris 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