Beat 99% Fast Java Solution O(h) Space with Explanation

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    The basic idea is to use stack to do in-order traversal. In the processing of traversal, keep comparing the current value with the previous value. Since each previous value should be less than the current value, once an exception is found, record the previous node as the First Mistaken Node and the current node as Second. If one more exceptions are found, override the current node to the Second Mistaken Node. Because if a series of mistaken nodes are found, the only possible way to correct them with one swap is to switch the head and tail node.

    public void recoverTree(TreeNode root) {
    	TreeNode pre = null, first = null, second = null;
    	Deque<TreeNode> stack = new LinkedList<TreeNode>();
    	while (root != null) {
    		root = root.left;
    	while (!stack.isEmpty()) {
    		TreeNode temp = stack.pop();
    		if (pre != null)
    			if (pre.val > temp.val) {
    				if (first == null)
    					first = pre;
    				second = temp;
    		pre = temp;
    		if (temp.right != null) {
    			temp = temp.right;
    			while (temp != null) {
    				temp = temp.left;
    	int temp = first.val;
    	first.val = second.val;
    	second.val = temp;

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