Description
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 1:
Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
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. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
Example 2:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 212 - 1].
-1000 <= Node.val <= 1000
Follow-up:
- You may only use constant extra space.
- The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.
Solutions
Solution 1: BFS
Use a queue for level order traversal, and each time you traverse a level, connect the nodes of the current level in order.
The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes in the binary tree.
PythonJavaC++GoTypeScript
"""
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: "Optional[Node]") -> "Optional[Node]":
if root is None:
return root
q = deque([root])
while q:
p = None
for _ in range(len(q)):
node = q.popleft()
if p:
p.next = node
p = node
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
return root(code-box)
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
if (root == null) {
return root;
}
Deque<Node> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
Node p = null;
for (int n = q.size(); n > 0; --n) {
Node node = q.poll();
if (p != null) {
p.next = node;
}
p = node;
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
}
return root;
}
}(code-box)
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
Node(int _val, Node* _left, Node* _right, Node* _next)
: val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
Node* connect(Node* root) {
if (!root) {
return root;
}
queue<Node*> q{{root}};
while (!q.empty()) {
Node* p = nullptr;
for (int n = q.size(); n; --n) {
Node* node = q.front();
q.pop();
if (p) {
p->next = node;
}
p = node;
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
}
return root;
}
};(code-box)
/**
* Definition for a Node.
* type Node struct {
* Val int
* Left *Node
* Right *Node
* Next *Node
* }
*/
func connect(root *Node) *Node {
if root == nil {
return root
}
q := []*Node{root}
for len(q) > 0 {
var p *Node
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
if p != nil {
p.Next = node
}
p = node
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
}
return root
}(code-box)
/**
* Definition for Node.
* class Node {
* val: number
* left: Node | null
* right: Node | null
* next: Node | null
* constructor(val?: number, left?: Node, right?: Node, next?: Node) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function connect(root: Node | null): Node | null {
if (root == null || root.left == null) {
return root;
}
const { left, right, next } = root;
left.next = right;
if (next != null) {
right.next = next.left;
}
connect(left);
connect(right);
return root;
}(code-box)
Solution 2: DFS
Use recursion for preorder traversal, and each time you traverse to a node, connect its left and right child nodes in order.
Specifically, we design a function dfs(left, right), which points the next pointer of the left node to the right node. In the function, we first check whether left and right are null. If both are not null, point left.next to right, and then recursively call dfs(left.left, left.right), dfs(left.right, right.left), dfs(right.left, right.right).
The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes in the binary tree.
PythonJavaC++GoTypeScript
"""
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution:
def connect(self, root: 'Optional[Node]') -> 'Optional[Node]':
def dfs(left, right):
if left is None or right is None:
return
left.next = right
dfs(left.left, left.right)
dfs(left.right, right.left)
dfs(right.left, right.right)
if root:
dfs(root.left, root.right)
return root(code-box)
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
if (root != null) {
dfs(root.left, root.right);
}
return root;
}
private void dfs(Node left, Node right) {
if (left == null || right == null) {
return;
}
left.next = right;
dfs(left.left, left.right);
dfs(left.right, right.left);
dfs(right.left, right.right);
}
}(code-box)
/*
// Definition for a Node.
class Node {
public:
int val;
Node* left;
Node* right;
Node* next;
Node() : val(0), left(NULL), right(NULL), next(NULL) {}
Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
Node(int _val, Node* _left, Node* _right, Node* _next)
: val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
Node* connect(Node* root) {
function<void(Node*, Node*)> dfs = [&](Node* left, Node* right) {
if (!left || !right) {
return;
}
left->next = right;
dfs(left->left, left->right);
dfs(left->right, right->left);
dfs(right->left, right->right);
};
if (root) {
dfs(root->left, root->right);
}
return root;
}
};(code-box)
/**
* Definition for a Node.
* type Node struct {
* Val int
* Left *Node
* Right *Node
* Next *Node
* }
*/
func connect(root *Node) *Node {
var dfs func(*Node, *Node)
dfs = func(left, right *Node) {
if left == nil || right == nil {
return
}
left.Next = right
dfs(left.Left, left.Right)
dfs(left.Right, right.Left)
dfs(right.Left, right.Right)
}
if root != nil {
dfs(root.Left, root.Right)
}
return root
}(code-box)
/**
* Definition for Node.
* class Node {
* val: number
* left: Node | null
* right: Node | null
* next: Node | null
* constructor(val?: number, left?: Node, right?: Node, next?: Node) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function connect(root: Node | null): Node | null {
if (root == null) {
return root;
}
const queue = [root];
while (queue.length !== 0) {
const n = queue.length;
let pre = null;
for (let i = 0; i < n; i++) {
const node = queue.shift();
node.next = pre;
pre = node;
const { left, right } = node;
left && queue.push(right, left);
}
}
return root;
}(code-box)
