In order traversal

99. Recover Binary Search Tree

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?

public class Solution {
    
    TreeNode firstElement = null;
    TreeNode secondElement = null;
    // The reason for this initialization is to avoid null pointer exception in the first comparison when prevElement has not been initialized
    TreeNode prevElement = new TreeNode(Integer.MIN_VALUE);
    
    public void recoverTree(TreeNode root) {
        
        // In order traversal to find the two elements
        traverse(root);
        
        // Swap the values of the two nodes
        int temp = firstElement.val;
        firstElement.val = secondElement.val;
        secondElement.val = temp;
    }
    
    private void traverse(TreeNode root) {
        
        if (root == null)
            return;
            
        traverse(root.left);
        
        // Start of "do some business", 
        // If first element has not been found, assign it to prevElement (refer to 6 in the example above)
        if (firstElement == null && prevElement.val >= root.val) {
            firstElement = prevElement;
        }
    
        // If first element is found, assign the second element to the root (refer to 2 in the example above)
        if (firstElement != null && prevElement.val >= root.val) {
            secondElement = root;
        }        
        prevElement = root;

        // End of "do some business"

        traverse(root.right);
}

注意：可以通过in order遍历来得到树的递增序列，然后再判断顺序是否正确。在in order中还可以找到前序元素，如上文中的prevElement