[leetcode] restore binary search tree

#LeetCode A daily topic 【 Special topic of binary tree 】

• Restore binary search tree
  https://leetcode-cn.com/problems/recover-binary-search-tree/
• analysis
  An important characteristic of binary search tree is that the result of its order traversal is ascending , The binary search tree is swapped for two nodes , It is equivalent to that two numbers are exchanged in the ascending array . How to find two numbers exchanged in an ascending array ？
  Suppose an array [1,2,3,4,5,6,7] Swap one of ,2、6 by [1,6,3,4,5,2,7],
  When the current element is larger than the value of the next element for the first time, it must be the first position of the error （ One of the elements being exchanged ）, Set to error1, Then the next step is to find error1 In the right place . The above example is 6>3,error1 by 6.
  Keep going back , For the first time, it is better than error1 Big elements , His last one was error2 The location of , That is, the second element that went wrong （ Two of the elements exchanged ）. The above example is 7>6,error2 by 7 Former one 2. Of course, if you don't find anything better than error1 Big elements , Then he is the biggest , It should be at the end of the array .
• Realization

public void recoverTree(TreeNode root) {
    
        TreeNode node = root;
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode error1 = null, error2 = null, prev = null;
        //  In the sequence traversal 
        while (node != null || !stack.isEmpty()) {
    
            while (node != null) {
    
                stack.push(node);
                node = node.left;
            }
            node = stack.pop();
            if (prev != null && error1 == null && node.val < prev.val) {
    
                //  Wrong page 1 Nodes , First appearance i Less than i-1,i-1 Must be the first wrong value 
                error1 = prev;
            }
            //  find error2 Where it should be , That is, the first occurrence greater than error1 The value of , The previous one should be error2 The location of 
            //  If not , It must be the last one 
            if (error1 != null && node.val > error1.val) {
    
                error2 = prev;
                break;
            }
            prev = node;
            node = node.right;
        }
        //  In exchange for 
        // error2 It's empty , Then the last one is error2
        if (error1 != null) {
    
            error2 = error2 == null ? prev : error2;
            int value = error1.val;
            error1.val = error2.val;
            error2.val = value;
        }
    }

LeetCode Time consuming ：1ms

• summary

1. Master the realization of middle order traversal , Including recursive implementation and non recursive implementation
2. The binary search tree is followed by an ascending array
3. After minimizing the problem , First, simulate more situations , ideas , Do it again