Binary Tree Level Order Traversal II
2018-01-29 23:01
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题目描述
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).Queue使用时要尽量避免Collection的add()和remove()方法,而是要**
使用offer()来加入元素,使用poll()来获取并移出元素。
它们的优点是通过返回值可以判断成功与否,add()和remove()方法在失败的时候会抛出异常。 如果要使用前端而不移出该元素,使用element()或者peek()方法。
解题思路: 层序遍历,使用队列来实现,最后结果将数组逆置即可
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { //层序遍历二叉树 List<List<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(); List<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; } }
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