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More:【目录】LeetCode Java实现
https://leetcode.com/problems/binary-tree-level-order-traversal/
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] ]
Using a queue.
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
LinkedList<TreeNode> queue = new LinkedList<>();
if(root!=null)
queue.offer(root);
while(!queue.isEmpty()){
int num = queue.size();
List<Integer> subList = new LinkedList<Integer>();
for(int i=0; i<num; i++){
TreeNode node = queue.poll();
subList.add(node.val);
if(node.left!=null)
queue.offer(node.left);
if(node.right!=null)
queue.offer(node.right);
}
wrapList.add(subList);
}
return wrapList;
}
Time complexity : O(n)
Space complexity : O(n)
More:【目录】LeetCode Java实现
【LeetCode】102. Binary Tree Level Order Traversal
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原文地址：https://www.cnblogs.com/yongh/p/11791511.html
