力扣0107——二叉树的层序遍历II

二叉树的层序遍历II

难度：中等

题目描述

给你二叉树的根节点 root ，返回其节点值 自底向上的层序遍历 。（即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

示例1

输入： inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]

示例2

输入： inorder = [-1], postorder = [-1]
输出：[-1]

题解

和 0102 题解题思路相同，不同的只不过是最后将List加入到List<List>中的时候变为Insert即可

想法代码

public class TreeNode
{public int val;public TreeNode left;public TreeNode right;public TreeNode(int val = 0, TreeNode left = null, TreeNode right = null){this.val = val;this.left = left;this.right = right;}
}class Solution
{public static void Main(string[] args){TreeNode root = new TreeNode(3){left = new TreeNode(9),right = new TreeNode(20){left = new TreeNode(15),right = new TreeNode(7)}};Solution solution = new Solution();IList<IList<int>> ans = solution.LevelOrderBottom(root);foreach (var a in ans){foreach (var b in a){Console.Write(b + " ");}Console.WriteLine();}}public IList<IList<int>> LevelOrderBottom(TreeNode root){IList<IList<int>> ans = new List<IList<int>>();if (root == null){return ans;}Queue<TreeNode> queue = new Queue<TreeNode>();queue.Enqueue(root);while (queue.Count > 0){IList<int> tmp = new List<int>();int count = queue.Count;for (int i = 0; i < count; i++){TreeNode treeNode = queue.Dequeue();tmp.Add(treeNode.val);if (treeNode.left != null){queue.Enqueue(treeNode.left);}if (treeNode.right != null){queue.Enqueue(treeNode.right);}}ans.Insert(0,tmp);}return ans;}
}