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# Tree

## 117. Populating Next Right Pointers in Each Node II

## 1. Question

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
```

## 2. Implementation

**(1) BFS**

```java
public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) {
            return;
        }

        Queue<TreeLinkNode> queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeLinkNode curNode = queue.remove();

                if (i == size - 1) {
                    curNode.next = null;
                }
                else {
                   curNode.next = queue.peek(); 
                }

                if (curNode.left != null) {
                    queue.add(curNode.left);
                }

                if (curNode.right != null) {
                    queue.add(curNode.right);
                }
            }
        }
    }
}
```

**(2) Iteration**

```java
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode preNode = null, curNode = root, leftMostNode = null;

        while (curNode != null) {
            while (curNode != null) {

                if (curNode.left != null) {
                    if (preNode != null) {
                        preNode.next = curNode.left;
                    }
                    else {
                        leftMostNode = curNode.left;
                    }
                    preNode = curNode.left;
                }

                if (curNode.right != null) {
                    if (preNode != null) {
                        preNode.next = curNode.right;
                    }
                    else {
                        leftMostNode = curNode.right;
                    }
                    preNode = curNode.right;
                }

                curNode = curNode.next;
            }
            curNode = leftMostNode;
            preNode = null;
            leftMostNode = null;
        }
    }
}
```

## 3. Time & Space Complexity

BFS: 时间复杂度: O(n), 空间复杂度: O(w), w是树中拥有最多node的一层中的node个数

Iteration: 时间复杂度: O(n), 空间复杂度: O(1)