Leetcode
Tree
Breath-first Search
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7]
,
3 / 9 20 / 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
分析¶
这道题目要求的层序遍历。题目这么长,其实想让我们写广度优先搜索。
public List<List<Integer>> levelOrder(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); } return levels; }
不过也可以用深度优先搜索写
public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<List<Integer>>(); levelHelper(res, root, 0); return res; } public void levelHelper(List<List<Integer>> res, TreeNode root, int height) { if (root == null) return; if (height == res.size()) { res.add(new LinkedList<Integer>()); } res.get(height).add(root.val); levelHelper(res, root.left, height+1); levelHelper(res, root.right, height+1); }
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