LeetCode:102. Binary Tree Level Order Traversal(二叉樹的層次遍歷)
阿新 • • 發佈:2018-12-21
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] ]
方法1:深度優先(遞迴演算法)
/** * 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>> 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); } }
時間複雜度:O(n)
空間複雜度:O(n)
方法2:一種非遞迴的方式,利用棧的形式
/** * 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>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<List<Integer>>(); Queue<TreeNode> queue=new LinkedList<>(); queue.add(root); while(!queue.isEmpty()){ List<Integer> list=new ArrayList(); int size=queue.size(); for(int i=0;i<size;i++){ TreeNode p=queue.poll(); list.add(p.val); if(p.left!=null){ queue.add(p.left); } if(p.right!=null){ queue.add(p.right); } } res.add(list); } return res; } }
時間複雜度:O(n)
空間複雜度:O(n)