Leetcode
Tree
Breath-first Search
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] ]
分析¶
这道题目是LeeCode 102. Binary Tree Level Order Traversal的扩展，唯一的区别是这道题目要求从底向上的遍历。所以非常直接的方法就是把LeetCode102的结果反转一下：
public List<List<Integer>> levelOrderBottom(TreeNode root) { List<List<Integer>> levels = new ArrayList<List<Integer>>(); if (root == null) return levels; Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { List<Integer> curLevel = new ArrayList<>(); int size = queue.size(); 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); curLevel.add(node.val); } levels.add(curLevel); } Collections.reverse(levels); return levels; }
或者使用链表，每次添加在链表首部
public List<List<Integer>> levelOrderBottom(TreeNode root) { LinkedList<List<Integer>> levels = new LinkedList<List<Integer>>(); if (root == null) return levels; Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { List<Integer> curLevel = new ArrayList<>(); int size = queue.size(); 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); curLevel.add(node.val); } levels.addFirst(curLevel); } return levels; }
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