leetcode 102. Binary Tree Level Order Traversal

2024年5月24日 197次阅读

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]
]

解题思路：
如果知道二叉树层序便利就知道用一个队列可以层序便利，那么如果把每层的节点装到一个list中，关键就在于用另一个队列装子节点。于是解法如下：

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> resultlist = new ArrayList<>();
        Queue queue1 = new LinkedList<Integer>();
        Queue queue2 = new LinkedList<Integer>();
        if(root==null)
            return resultlist;
        queue1.add(root);
        List<Integer> templist = new ArrayList<Integer>();
        templist.add(root.val);
        resultlist.add(templist);
        while(!(queue1.peek()==null&&queue2.peek()==null)){
            templist = new ArrayList<Integer>();
            while(queue1.peek()!=null){
                TreeNode temp = (TreeNode)queue1.poll();
                if(temp.left!=null){
                    queue2.add(temp.left);
                    templist.add(temp.left.val);
                }
                if(temp.right!=null){
                    queue2.add(temp.right);
                    templist.add(temp.right.val);
                }
            }
            if(templist.size()!=0&&templist!=null){
                resultlist.add(templist);
            }
            while(queue2.peek()!=null){
                queue1.add(queue2.poll());
            }
        }
        return resultlist;

    }
}
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