LeetCode 102. 二叉树的层序遍历

102. 二叉树的层序遍历

Solution

思路：搞一个深度，然后放到对应的层次里。这里开List嵌套的时候有点坑，不过根据list的容量大小和层次的高度关系，来进行创建。这里的NewNode不能共用。题解代码是 用for循环搞出来那一层的。普通的BFS是弹出一个。见下面代码二。
更新：LeetCode 429 N叉树的层序遍历。代码三

///代码一
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null) return new ArrayList<>();
        List<List<Integer>> lists = new ArrayList<>();
        Queue<NewNode> queue = new ArrayDeque<>();
        NewNode newNode = new NewNode();
        newNode.node = root;
        newNode.deep = 0;
        queue.add(newNode);
        while (!queue.isEmpty()) {
            NewNode t = queue.remove();
            if (lists.size() <= t.deep) {
                lists.add(new ArrayList<>());
            }
            lists.get(t.deep).add(t.node.val);
            if (t.node.left != null) {
                NewNode l = new NewNode();
                l.node = t.node.left;
                l.deep = t.deep + 1;
                queue.add(l);
            }
            if (t.node.right != null) {
                NewNode r = new NewNode();
                r.node = t.node.right;
                r.deep = t.deep + 1;
                queue.add(r);
            }
        }
        return lists;
    }
}

class NewNode {
    TreeNode node;
    int deep;
}

//代码二
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> lists = new ArrayList<>();
        if (root == null) return lists;
        Queue<TreeNode> queue = new ArrayDeque<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            List<Integer> level = new ArrayList<>();
            int sz = queue.size();
            for (int i = 0; i < sz; i++) {
                TreeNode node = queue.remove();
                level.add(node.val);
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
            lists.add(level);
        }
        return lists;
    }
}

//代码三
class Node {
    public int val;
    public List<Node> children;

    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>> ans = new ArrayList<>();
        if (root == null) return ans;
        Deque<Node> queue = new ArrayDeque<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int sz = queue.size();
            List<Integer> level = new ArrayList<>();
            for (int i = 0; i < sz; i++) {
                Node child = queue.remove();
                level.add(child.val);
                for (Node node : child.children) {
                    queue.add(node);
                }
            }
            ans.add(level);
        }
        return ans;
    }
}