二叉树的层序遍历 II

news2024/11/15 9:29:22

题目链接

二叉树的层序遍历 II

题目描述

注意点

树中节点数目在范围 [0, 2000] 内
-1000 <= Node.val <= 1000

解答思路

根据队列先进先出的特点层序遍历所有的节点（从左到右），又因为需要自底向上的输出层序遍历的结果，所以可以将每一次遍历的结果存储到栈中，根据栈先进后出的特点将结果输出

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Deque<List<Integer>> stack = new ArrayDeque<>();
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>(size);
            stack.push(list);
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
                list.add(node.val);
            }
        }
        while (!stack.isEmpty()) {
            res.add(stack.pop());
        }
        return res;
    }
}