题目

代码

/**

• Definition for a binary tree node.

• public class TreeNode {

• int val;
  

• TreeNode left;
  

• TreeNode right;
  

• TreeNode() {}
  

• TreeNode(int val) { this.val = val; }
  

• TreeNode(int val, TreeNode left, TreeNode right) {
  

•     this.val = val;
  

•     this.left = left;
  

•     this.right = right;
  

• }
  

• }
  */
  class Solution {
  public int[] decorateRecord(TreeNode root) {
  if(root == null){
  return new int[0];
  }

   Queue<TreeNode> queue = new LinkedList<>();
   List<Integer> res = new ArrayList<>();
  
   queue.add(root);
   while(!queue.isEmpty()){
       TreeNode t = queue.poll();
       res.add(t.val);
       if(t.left != null) queue.add(t.left);
       if(t.right != null) queue.add(t.right);
   }
  
   int[] arr = new int[res.size()];
   for(int i = 0; i < res.size(); i++){
       arr[i] = res.get(i);
   }
  
   return arr;
  

  }
  }
  时间复杂度：O(n)
  额外空间复杂度：容器里最对存放 1/2 的节点，故为 O(n)