117. Populating Next Pointers in Each Node II

Question

Given a binary tree

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.

Solution

BFS搜索，从右至左将每层节点放入队列。
用一个size记录该层级放入节点的数量，每次循环将size-1，当size归零时放入新一层级的节点。

Code

/*
// 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;
        
        Queue<Node> q = new LinkedList();
        q.add(root);
        
        while(!q.isEmpty()){
            int size = q.size();
            Node next = null;
            while(size > 0){
                size--;
                Node curr = q.poll();
                curr.next = next;
                next = curr;
                if(curr.right != null) q.add(curr.right);
                if(curr.left != null) q.add(curr.left);
            }
        }
        return root;
    }
}