Day13 | 二叉树part 01- 102.二叉树的层序遍历(opens new window) 两个数组一个队列

- 102.二叉树的层序遍历(opens new window)

两个数组

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null) return List.of();
        List<List<Integer>> ans = new ArrayList<>();
        List<TreeNode> cur = List.of(root);
        while (!cur.isEmpty()) {
            List<TreeNode> nxt = new ArrayList<>();
            List<Integer> vals = new ArrayList<>(cur.size()); // 预分配空间
            for (TreeNode node : cur) {
                vals.add(node.val);
                if (node.left != null)  nxt.add(node.left);
                if (node.right != null) nxt.add(node.right);
            }
            cur = nxt;
            ans.add(vals);
        }
        return ans;
    }
}

一个队列

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null) return List.of();
        List<List<Integer>> ans = new ArrayList<>();
        Queue<TreeNode> q = new ArrayDeque<>();
        q.add(root);
        while (!q.isEmpty()) {
            int n = q.size();
            List<Integer> vals = new ArrayList<>(n); // 预分配空间
            while (n-- > 0) {
                TreeNode node = q.poll();
                vals.add(node.val);
                if (node.left != null)  q.add(node.left);
                if (node.right != null) q.add(node.right);
            }
            ans.add(vals);
        }
        return ans;
    }
}

- 107.二叉树的层次遍历II(opens new window)

代码同上，两种方法，反转添加如下语句：

Collections.reverse(ans);

- 199.二叉树的右视图(opens new window)

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        dfs(root, 0, ans);
        return ans;
    }
    private void dfs(TreeNode root, int depth, List<Integer> ans) {
        if (root == null) {
            return;
        }
        if (depth == ans.size()) {
            ans.add(root.val);
        }
        dfs(root.right, depth + 1, ans);
        dfs(root.left, depth + 1, ans);
    }
}

这个题，挺难理解。