117. Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

---

同１１６题类似，用中序遍历的话，需要Ｏ（n)的时间复杂度。
通过前一行的ｎｅｘｔ，来找下一行的ｎｅｘｔ。用两个指针prevLevelStart和prev, 分别指向前一行有孩子的第一个节点，和用来连接当前行时、前一行的辅助节点。

与116相比，只是updatePrevLevelStart和ｕｐｄａｔｅＰｒｅｖ的规则变了。

对１１６由于是perfect binary tree:

     prevLevelStart = prevLevelStart.left

     prev = prev.next

对１１７题，要跳过空节点。 ３０％

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode prevLevelStart = root, prev, cur;
        while (prevLevelStart != null) {
            prev = prevLevelStart;
            if (prev.left != null) {
                cur = prev.left;
            } else {
                cur = prev.right;
                prev = updatePrev(prev);
            }
            while (prev != null) {
                if (cur == prev.left) {
                    if (prev.right != null) {
                        cur.next = prev.right;
                        cur = cur.next;
                    }
                    prev = updatePrev(prev);   
                } else {
                    if (prev.left != null) {
                        cur.next = prev.left;
                        cur = cur.next;
                    } else {
                        cur.next = prev.right;
                        cur = cur.next;
                        prev = updatePrev(prev);
                    }
                }
            }
            prevLevelStart = updatePrevLevelStart(prevLevelStart);
        }
    }
    public TreeLinkNode updatePrev(TreeLinkNode prev) {
        prev = prev.next;
        while (prev != null && prev.left == null && prev.right == null) {
            prev = prev.next;
        }
        return prev;
    }
    public TreeLinkNode updatePrevLevelStart(TreeLinkNode prevLevelStart) {
        while (prevLevelStart != null) {
            if (prevLevelStart.left != null && (prevLevelStart.left.left != null || prevLevelStart.left.right != null)) {
                return prevLevelStart.left;
            } else if (prevLevelStart.right != null && (prevLevelStart.right.left != null || prevLevelStart.right.right != null)) {
                return prevLevelStart.right;
            } else {
                prevLevelStart = prevLevelStart.next;
            }
        }
        return null;
    }
}