Recover Binary Search Tree--leetcode難題講解

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?

https://leetcode.com/problems/recover-binary-search-tree/

二叉排序樹中有兩個節點被交換了，要求把樹恢復成二叉排序樹。

最簡單的辦法，中序遍歷二叉樹生成序列，而後對序列中排序錯誤的進行調整。最後再進行一次賦值操做，但這個不符合空間複雜度要求。

須要用兩個指針記錄錯誤的那兩個元素，而後進行交換。

怎樣找出錯誤的元素？遍歷二叉排序樹，正確時應該是從小到大，若是出現以前遍歷的節點比當前大，則說明出現錯誤。因此咱們須要一個pre指針來指向以前通過的節點。

若是隻有一處不符合從小到大，則只用交換這一個地方。第二個指針記錄第二個異常點。

 

 Github repository: https://github.com/huashiyiqike/myleetcode 

//JAVA CODE：
public class Solution {
    TreeNode first = null, second = null, pre = null;
     //first larger than follow, second smaller than pre
    public void helper(TreeNode root){
        if(root.left != null) helper(root.left);
        if(pre != null && root.val < pre.val){
            if(first == null)
                first = pre;
            second = root;
        }
        pre = root;
        if(root.right != null) helper(root.right);
    }
    public void recoverTree(TreeNode root) {
        helper(root);
        int tmp = first.val;
        first.val = second.val;
        second.val = tmp;
    }
}

 

//C++ CODE：
#include <iostream>
#include <cstdlib>
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


class Solution {
    TreeNode *first = NULL, *second = NULL, *last = NULL;
public:
    void inorder(TreeNode* root){
        if(root->left != NULL){
            inorder(root->left);
        }
        if(last != NULL && root->val < last->val){
            if(first == NULL) first = last;
            second = root;
        }
        last = root;
        if(root->right != NULL) inorder(root->right);
    }
    void recoverTree(TreeNode* root) {
        if(root == NULL) return;
        inorder(root);
        if(first && second)
            std::swap(first->val, second->val);
    }
};

#PYTHON CODE：
class Solution:
    def inorderTravel(self, root, last):
        if root.left:
            last = self.inorderTravel(root.left, last)
        if last and last.val > root.val:
            if self.first is None:
                self.first = last
            self.second = root
        last = root
        if root.right:
            last = self.inorderTravel(root.right, last)
        return last

    def recoverTree(self, root):
        if root is None:
            return
        self.first, self.second = None, None
        self.inorderTravel(root, None)
        self.first.val, self.second.val = self.second.val, self.first.val