Recover Binary Search Tree

Recover Binary Search Tree

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Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:

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

confused what 

"{1,#,2,3}"

 means? >
read more on how binary tree is serialized on OJ.

这题关键在于inorder traversal. 首先需要3个全局指针pre,first,second。 pre始终指向traversal当前node的前一个，first是第一个错误的点，second是第二个错误的点。

注意点：两个错误点可能是相邻的，也可能是不相邻的。当第一次发现错误点的时候， first = pre, second = root, 把两个错误点都设了，防止之后没有错误点了，那么就是这两个了。

如果之后还有错误点，只更新 second 即可。

static TreeNode pre = null, first = null, second = null;
public static void recoverTree(TreeNode root) {
recoverTreeHelper(root);
if (first != null && second != null) {
int temp = first.val;
first.val = second.val;
second.val = temp;
}
}

public static void recoverTreeHelper(TreeNode root) {
if (root == null) return;
recoverTreeHelper(root.left);
if (pre == null) pre = root;
else if (root.val < pre.val) {
if (first == null) {
first = pre;
}
second = root;
}
pre = root;
recoverTreeHelper(root.right);
}