116. Populating Next Right Pointers in Each Node #

Problem #

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.

Follow up:

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

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.

Constraints:

• The number of nodes in the given tree is less than 4096.
• 1000 <= node.val <= 1000

Problem Summary #

Given a perfect binary tree, where all leaves are on the same level and every parent has two children. The binary tree is defined as follows:

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

Populate each next pointer so that it points to its next right node. If there is no next right node, set the next pointer to NULL. Initially, all next pointers are set to NULL.

Solution Approach #

• Essentially, this is a level-order traversal of a binary tree. Based on breadth-first search, put the nodes of each level into a queue, and traverse the queue to connect them.

Code #

package leetcode

type Node struct {
	Val   int
	Left  *Node
	Right *Node
	Next  *Node
}

//Solution 1: Iteration
func connect(root *Node) *Node {
	if root == nil {
		return root
	}
	q := []*Node{root}
	for len(q) > 0 {
		var p []*Node
		// Traverse all nodes at this level
		for i, node := range q {
			if i+1 < len(q) {
				node.Next = q[i+1]
			}
			if node.Left != nil {
				p = append(p, node.Left)
			}
			if node.Right != nil {
				p = append(p, node.Right)
			}
		}
		q = p
	}
	return root
}

// Solution 2: Recursion
func connect2(root *Node) *Node {
	if root == nil {
		return nil
	}
	connectTwoNode(root.Left, root.Right)
	return root
}

func connectTwoNode(node1, node2 *Node) {
	if node1 == nil || node2 == nil {
		return
	}
	node1.Next = node2
	connectTwoNode(node1.Left, node1.Right)
	connectTwoNode(node2.Left, node2.Right)
	connectTwoNode(node1.Right, node2.Left)
}