Description

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

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

1 / 3 2

Output: [3,1,null,null,2]

3 / 1 2 Example 2:

Input: [3,1,4,null,null,2]

3 / 1 4 / 2

Output: [2,1,4,null,null,3]

2 / 1 4 / 3 Follow up:

A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

Solution

二叉搜尋樹中兩個節點互換了，恢復。

In-order traversal the tree. find the first element bigger than the node after. and a node smaller than the node ahead of it. Swap these two node.

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 

class Solution {
    TreeNode first = null;
    TreeNode second = null;
    TreeNode prev = new TreeNode(Integer.MIN_VALUE);
    public void recoverTree(TreeNode root) {
        inOrder(root);
        
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
    }
    
    private
 
 void inOrder(TreeNode node){
        if (node == null){
            return;
        }
        inOrder(node.left);
        if (first == null && prev.val > node.val){
            first = prev;
        }
        
        if (first != null && prev.val > node.val){
            second = node;
        }
            
        prev = node;
        inOrder(node.right);
    }
}

Time Complexity: O(n) Space Complexity: O(1)

Review