【LeetCode每日打卡】102. 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]
]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-level-order-traversal
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层序遍历Java实现

我这个也不是层序遍历，而是中序遍历，但是插入到数组中。

本质上还是DFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<List<Integer>> ans;
    public List<List<Integer>> levelOrder(TreeNode root) {
        ans = new ArrayList();
        levelO(ans,root,0);
        return ans;
    }
    public void levelO(List<List<Integer>>ans,TreeNode root,int high){
        if(root == null){
            return;
        }
        if(ans.size()<=high){
            ans.add(new ArrayList());
        }
        ans.get(high).add(root.val);
        levelO(ans,root.left,high+1);
        levelO(ans,root.right,high+1);
    }
}

我认为真正的层序遍历 还是BFS 

list暂存一层的节点，存入result之后再遍历下一层。

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
		if (root == null) {
			return result;
		}

		LinkedList<TreeNode> list = new LinkedList<>();
		list.add(root);

		while (!list.isEmpty()) {
			int size = list.size();
			List<Integer> levelVal = new ArrayList<>();
			for (int i = 0; i < size; i++) {
				TreeNode treeNode = list.pollFirst();
				if (treeNode != null) {
					levelVal.add(treeNode.val);
					if (treeNode.left != null) {
						list.add(treeNode.left);
					}
					if (treeNode.right != null) {
						list.add(treeNode.right);
					}
				}
			}
			result.add(levelVal);
		}

		return result;
    }
}