Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Ref:http://fisherlei.blogspot.com/2012/12/leetcode-recover-binary-search-tree.html

O(1)的解法就是
Inorder traverse, keep the previous tree node,
Find first misplaced node by
if ( current.val < prev.val )
Node first = prev;

Find second by
if ( current.val < prev.val )
Node second = current;

After traversal, swap the values of first and second node. Only need two pointers, prev and current node. O(1) space.

但是这个解法的前提是Traverse Tree without Stack. 中序遍历如何才能不使用栈。这里就要引入一个概念， Threaded Binary Tree。So, we first create links to Inorder successor and print the data using these links, and finally revert the changes to restore original tree.

1. Initialize current as root
2. While current is not NULLIf current does not have left childa) Print current’s datab) Go to the right, i.e., current = current->rightElsea) Make current as right child of the rightmost node in current's left subtreeb) Go to this left child, i.e., current = current->left

public class Solution {public void recoverTree(TreeNode root) {TreeNode f1 = null, f2 = null;TreeNode current, pre, parent = null;if(root == null)return;// 用来记录是第几次找到bad node, 因为第一次找到的bad node为cur.pre，第二次找到bad node是curboolean found = false;current = root;while(current != null){                if(current.left == null){if(parent != null && parent.val > current.val){if(!found){f1 = parent;found = true;}f2 = current;}parent = current;current = current.right;     }   else{/* Find the inorder predecessor of current */pre = current.left;while(pre.right != null && pre.right != current)pre = pre.right;/* Make current as right child of its inorder predecessor */if(pre.right == null){pre.right = current;current = current.left;}/* Revert the changes made in if part to restore the originaltree i.e., fix the right child of predecssor */  else{// 注意跳进这里的原因是，当把root 设为左子树的最右节点时，root.left 仍然为左子树的根节点，相当于生成了一个环pre.right = null;if(parent.val > current.val){if(!found){f1 = parent;       found = true;}// f2 不写在else里的原因是 有可能是相邻的两个元素交换了f2 = current;}parent = current;current = current.right;     } /* End of if condition pre->right == NULL */} /* End of if condition current->left == NULL*/} /* End of while */if(f1 != null && f2 != null){int tmp = f1.val;f1.val = f2.val;f2.val = tmp;}}
}

/*** Definition for binary tree* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Solution {public static void recoverTree(TreeNode root) {ArrayList<TreeNode> inorderList = new ArrayList<TreeNode>();inorder(root, inorderList);int first= -1, second  = -1; boolean secondNumber = false;for(int i = 1; i < inorderList.size();i++){if (inorderList.get(i-1).val > inorderList.get(i).val){if(!secondNumber){first = i-1;second = i;secondNumber = true;}else{second = i;}}}swap(inorderList.get(first), inorderList.get(second));}private static void inorder(TreeNode root, ArrayList<TreeNode> inorderList){if(root != null){inorder(root.left, inorderList);inorderList.add(root);inorder(root.right, inorderList);}}private static void swap(TreeNode i, TreeNode j){int tmp = i.val;i.val = j.val;j.val = tmp;}}

转载于:https://www.cnblogs.com/RazerLu/p/3557776.html

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