[LeetCode]--107. Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

For example: Given binary tree [3,9,20,null,null,15,7],

3 / \ 9 20 / \ 15 7

return its bottom-up level order traversal as:

[ [15,7], [9,20], [3] ]

想着用之前顺序的方法，但是List在add的时候用栈来存放，然后再pop出来就逆序了。

public List<List<Integer>> levelOrder1(TreeNode root) { List<List<Integer>> list = new ArrayList<List<Integer>>(); if (root == null) return list; Queue<TreeNode> queue = new LinkedList<TreeNode>(); Stack<List<Integer>> stack = new Stack<List<Integer>>(); queue.add(root); while (!queue.isEmpty()) { List<Integer> list1 = new ArrayList<Integer>(); int size = queue.size(); for (int i = 0; i < size; i++) { TreeNode t1 = queue.poll(); list1.add(t1.val); if (t1.left != null) queue.add(t1.left); if (t1.right != null) queue.add(t1.right); } //list.add(list1); stack.push(list1); } while (!stack.isEmpty()) { list.add(stack.pop()); } return list; }

也是一次就通过了，再去看看别人的代码。

public class Solution { /** * @param root: The root of binary tree. * @return: buttom-up level order a list of lists of integer */ public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<>(); if (root == null) { return result; } Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { int size = queue.size(); ArrayList<Integer> level = new ArrayList<>(); for (int i = 0; i < size; i++) { TreeNode head = queue.poll(); level.add(head.val); if (head.left != null) { queue.offer(head.left); } if (head.right != null) { queue.offer(head.right); } } result.add(level); } Collections.reverse(result); return result; } }

最后用系统自带的reverse方法逆转也是可行的。