二叉树的层次遍历

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        //层序遍历，保存右边的
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            int size = queue.size();
            for(int i = 0; i < size; i++){
                TreeNode node = queue.poll();
                if(i == size - 1){
                    res.add(node.val);
                }
                if(node.left != null){
                    queue.offer(node.left);
                }
                if(node.right != null){
                    queue.offer(node.right);
                }
            }
        }
        return res;
    }
}

优化: 每层右孩子先入队。

if(node.right != null){

queue.offer(node.right);

}

if(node.left != null){

queue.offer(node.left);

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }

        //小优化：每层右孩子先入队。
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while(!queue.isEmpty()){
            int size = queue.size();
            for(int i = 0; i < size; i++){
                TreeNode node = queue.poll();
                if(i == 0){ //小优化：每层右孩子先入队
                    res.add(node.val);
                }
                
                if(node.right != null){
                    queue.offer(node.right);
                }
                if(node.left != null){
                    queue.offer(node.left);
                }
            }

        }
        return res;

    }
}

时间上优化了一点点

429.N叉树的层序遍历

一样的，把左右孩子换成遍历数组即可

if(children != null){

for (Node child : children) {

if (child != null) {

que.offerLast(child);

}

}

}

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;
//三个构造器
    public Node() {}
    public Node(int _val) {
        val = _val;
    }
    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> list = new ArrayList<>();
        Deque<Node> que = new LinkedList<>();
        if (root == null) {
            return list;
        }
        que.offer(root);
        while (!que.isEmpty()) {
            int levelSize = que.size();
            List<Integer> levelList = new ArrayList<>();
            for (int i = 0; i < levelSize; i++) {
                Node poll = que.pollFirst();
                levelList.add(poll.val);
                List<Node> children = poll.children;
                if(children != null){
                    for (Node child : children) {
                        if (child != null) {
                            que.offerLast(child);
                        }
                    }
                }
                
            }
            list.add(levelList);
        }
        return list;
    
    }
}