# 116. Populating Next Right Pointers in Each Node (Medium)

https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

```struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
```

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to `NULL`.

Initially, all next pointers are set to `NULL`.

Example:

```Input: {"\$id":"1","left":{"\$id":"2","left":{"\$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"\$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"\$id":"5","left":{"\$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"\$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"\$id":"1","left":{"\$id":"2","left":{"\$id":"3","left":null,"next":{"\$id":"4","left":null,"next":{"\$id":"5","left":null,"next":{"\$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"\$id":"7","left":{"\$ref":"5"},"next":null,"right":{"\$ref":"6"},"val":3},"right":{"\$ref":"4"},"val":2},"next":null,"right":{"\$ref":"7"},"val":1}

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.
```

Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

## Solutions

#### 1.

``````class Solution {

// Solution meets the requirements perfectly.

public Node connect(Node root) {
// return null if root empty
if (root == null) {
return root;
}

// link root.left  to root.right if root.left exists
if (root.left != null) {
root.left.next = root.right;
}

// link root.right to father's next's left child
if (root.right != null && root.next != null && root.next.left != null) {
root.right.next = root.next.left;
}

// then visit root.left and connect
connect(root.left);

// visit root.right and connect
connect(root.right);

return root;
}
}
``````

### 2.

``````class Solution {

// Solution violates the constraints. Ask for extra memory space.

public Node connect(Node root) {
if (root == null) {
return root;
}

// used to keep nodes for each level
Map<Integer, List<Node>> levelMap = new HashMap<>();

// populate nodes to same level
populateLayer(root, 0, levelMap);

// start to link node of each level
for (Integer k : levelMap.keySet()) {

// Nodes kept in list are arranged as the order from left to right.
List<Node> nodes = levelMap.get(k);

// since rightmost node has no siblings, ignore it.
for (int i = 0; i < nodes.size() - 1; i++) {
nodes.get(i).next = nodes.get(i + 1);
}
}

return root;
}

private void populateLayer(Node root, int level, Map<Integer, List<Node>> levelMap) {
if (root == null) {
return;
}

if (!levelMap.containsKey(level)) {
levelMap.put(level, new ArrayList<>());
}

populateLayer(root.left, level + 1, levelMap);
populateLayer(root.right, level + 1, levelMap);
}
}
``````

## Incorrect Solutions

``````class Solution {

// Incorrect solution

public Node connect(Node root) {
// return null if root empty
if (root == null) {
return root;
}

// link left child to right child if left child exists
if (root.left != null) {
root.left.next = root.right;
}

// then visit root.left and connect
connect(root.left);

// visit root.right and connect
connect(root.right);

return root;
}
}
``````