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

My code:

public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode level_start = root;
        while(level_start!=null){
            TreeLinkNode cur = level_start;
            while(cur!=null){
                TreeLinkNode temp = cur.next;
                while(temp!=null){
                    if(temp.left!=null) {
                        temp=temp.left;
                        break;
                    }
                    else if(temp.right!=null) {
                        temp=temp.right;
                        break;
                    }
                    else temp=temp.next;
                }
                if(cur.right!=null){
                    cur.right.next= temp;
                    if(cur.left!=null){
                        cur.left.next=cur.right;
                    }
                }
                else{
                    if(cur.left!=null)
                        cur.left.next=temp;
                }
                cur=cur.next;
            }
            while(level_start!=null){
                if(level_start.left!=null) {
                    level_start=level_start.left;
                    break;
                }
                else if(level_start.right!=null){
                    level_start=level_start.right;
                    break;
                }
                else level_start=level_start.next;
            }
        }
    }
}

总结：层次遍历

public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode head = null; //head of the next level
        TreeLinkNode prev = null; //the leading node on the next level
        TreeLinkNode cur = root;  //current node of current level

        while (cur != null) {
            
            while (cur != null) { //iterate on the current level
                //left child
                if (cur.left != null) {
                    if (prev != null) {
                        prev.next = cur.left;
                    } else {
                        head = cur.left;
                    }
                    prev = cur.left;
                }
                //right child
                if (cur.right != null) {
                    if (prev != null) {
                        prev.next = cur.right;
                    } else {
                        head = cur.right;
                    }
                    prev = cur.right;
                }
                //move to next node
                cur = cur.next;
            }
            
            //move to next level
            cur = head;
            head = null;
            prev = null;
        }
    }
}

总结：省去了while循环，把对当前cur的搜索，改为对cur的child的搜索。