LeetCode - Populating Next Right Pointers in Each Node
Problem statement
We 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.
Problem statement taken from: https://leetcode.com/problems/populating-next-right-pointers-in-each-node
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), our 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:
- We may only use constant extra space.
- The recursive approach is fine. We may assume implicit stack space does not count as extra space for this problem.
Explanation
If we see the problem statement, the tree is a perfect binary tree. We see in the example, the next right pointer for each node is at one level. We can do a level order traversal (as per this blog post) and update the next right pointer for each node.
Let's check the algorithm.
- if root == NULL
- return root
- initialize queue<Node*> q
- q.push(root)
- initialize Node* node
initialize i, size
- loop while !q.empty()
- set size = q.size()
- loop for i = 0; i < size; i++
- set node = q.front()
- if node->left
- q.push(node->left)
- if node->right
- q.push(node->right)
- q.pop()
- if i < size - 1
- node->next = q.front()
- return root
C++ solution
class Solution {
public:
Node* connect(Node* root) {
if(root == NULL)
return root;
queue<Node*> q;
q.push(root);
Node* node;
int i, size;
while(!q.empty()){
size = q.size();
for(i = 0; i < size; i++){
node = q.front();
if(node->left)
q.push(node->left);
if(node->right)
q.push(node->right);
q.pop();
if(i < size - 1){
node->next = q.front();
}
}
}
return root;
}
};Golang solution
func connect(root *Node) *Node {
if root == nil {
return root
}
queue := []*Node{root}
for len(queue) != 0 {
tmp := []int{}
size := len(queue)
for i := 0; i < size; i++ {
node := queue[0]
if queue[0] != nil {
tmp = append(tmp, queue[0].Val)
queue = append(queue, queue[0].Left)
queue = append(queue, queue[0].Right)
}
queue = queue[1:]
if i < size - 1 && queue[0] != nil {
node.Next = queue[0]
}
}
}
return root
}Javascript solution
var connect = function(root) {
let queue = [];
if(root)
queue.push(root);
while(queue.length > 0) {
tmp = [];
let len = queue.length;
for (let i = 0; i < len; i++) {
let node = queue.shift();
tmp.push(node.val);
if(node.left) {
queue.push(node.left);
}
if(node.right) {
queue.push(node.right);
}
if( i < len - 1 ) {
node.next = queue[0];
}
}
}
return root;
};Dry Run
Let's dry-run our algorithm to see how the solution works.
Input: root = [1, 2, 3, 4, 5, 6, 7]
Step 1: vector<vector<int>> result
int size, i
Step 2: queue<Node*> q
q.push(root)
q = [1]
Step 3: Node* node
int i, size
Step 4: loop while !q.empty()
q = [1]
q.empty() = false
!false = true
vector<int> tmp
size = q.size()
= 1
for(i = 0; i < 1; i++)
- 0 < 1
- true
node = q.front()
node = 1
if node->left
- node->left = 2
- q.push(node->left)
- q = [1, 2]
if node->right
- node->right = 3
- q.push(node->right)
- q = [1, 2, 3]
q.pop()
q = [2, 3]
if i < size - 1
- 0 < 1 - 1
- 0 < 0
- false
Step 5: loop while !q.empty()
q = [2, 3]
q.empty() = false
!false = true
vector<int> tmp
size = q.size()
= 2
for(i = 0; i < 2; i++)
- 0 < 2
- true
node = q.front()
node = 2
if node->left
- node->left = 4
- q.push(node->left)
- q = [2, 3, 4]
if node->right
- node->right = 5
- q.push(node->right)
- q = [2, 3, 4, 5]
q.pop()
q = [3, 4, 5]
if i < size - 1
- 0 < 2 - 1
- 0 < 1
- true
- node.next = q.front()
node = 2
q.front = 3
i++
i = 1
for(i < 2)
- 1 < 2
- true
node = q.front()
node = 3
if node->left
- node->left = 6
- q.push(node->left)
- q = [3, 4, 5, 6]
if node->right
- node->right = 7
- q.push(node->right)
- q = [3, 4, 5, 6, 7]
q.pop()
q = [4, 5, 6, 7]
if i < size - 1
- 1 < 2 - 1
- 1 < 1
- false
for(i < 2)
- 2 < 2
- false
Step 6: loop while !q.empty()
q = [4, 5, 6, 7]
q.empty() = false
!false = true
vector<int> tmp
size = q.size()
= 4
for(i = 0; i < 4; i++)
- 0 < 4
- true
node = q.front()
node = 4
if node->left
node->left is nil so false
if node->right
node->right is nil so false
q.pop()
q = [5, 6, 7]
if i < size - 1
- 0 < 4 - 1
- 0 < 3
- true
- node->next = q.front()
node = 4
q.front() = 5
i++
i = 1
for(i < 4)
- 1 < 4
- true
node = q.front()
node = 5
if node->left
node->left is nil so false
if node->right
node->right is nil so false
q.pop()
q = [6, 7]
if i < size - 1
- 1 < 4 - 1
- 1 < 3
- true
- node->next = q.front()
node = 5
q.front() = 6
i++
i = 2
for(i < 4)
- 2 < 4
- true
node = q.front()
node = 6
if node->left
node->left is nil so false
if node->right
node->right is nil so false
q.pop()
q = [7]
if i < size - 1
- 2 < 4 - 1
- 2 < 3
- true
- node->next = q.front()
node = 6
q.front() = 7
i++
i = 3
for(i < 4)
- 3 < 4
- true
node = q.front()
node = 7
if node->left
node->left is nil so false
if node->right
node->right is nil so false
q.pop()
q = []
if i < size - 1
- 3 < 4 - 1
- 3 < 3
- false
i++
i = 4
for(i < 4)
- 4 < 4
- false
Step 7: return root
So we return the answer as
[1, #, 2, 3, #, 4, 5, 6, 7, #]