0117. Populating Next Right Pointers in Each Node II #

题目 #

给定一个二叉树

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

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

思路 #

层序遍历

代码 #

层序遍历 #

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 null;
        
        Queue<Node> queue = new LinkedList<>();
        queue.offer(root);
        while (queue.size() != 0) {
            for (int i = 0, layerSize = queue.size(); i < layerSize; i++) {
                Node node = queue.poll();
                if (i != layerSize - 1) node.next = queue.peek();
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
        }
        
        return root;
    }
}