Question
Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).
Solution
BFS搜索,层序遍历,记录每个层级的节点数量size。
在遍历时挤出size个节点,并将其子节点加入队列。
Code
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class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> ret = new ArrayList<>(); if(root == null) return ret; Queue<TreeNode> q = new LinkedList<>(); q.offer(root);
int size = 1; while(!q.isEmpty()){ List<Integer> list = new ArrayList<>(); TreeNode curr = null; for(int i = 0; i < size; i++){ curr = q.poll(); if(curr.left != null) q.offer(curr.left); if(curr.right != null) q.offer(curr.right); list.add(curr.val); } ret.add(list); size = q.size(); } return ret; } }
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