python3 解法, 执行用时: 28 ms, 内存消耗: 15.2 MB, 提交时间: 2022-07-12 10:03:37

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if root is None: return []
        queue = [root]
        ans = []
        while len(queue) != 0:
            tmp = []
            cnt = len(queue)
            for i in range(cnt):
                node = queue[0]
                queue = queue[1:]
                tmp.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            ans.append(tmp)
        return ans

golang 解法, 执行用时: 0 ms, 内存消耗: 2.8 MB, 提交时间: 2021-07-23 10:09:17

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrder(root *TreeNode) (ans [][]int) {
    if root == nil {
        return
    }
    queue := []*TreeNode{root}

    for len(queue) != 0 {
        tmp := []int{}
        cnt := len(queue)
        for i := 0; i < cnt; i++ {
            node := queue[0]
            queue = queue[1:]
            tmp = append(tmp, node.Val)
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
        }
        ans = append(ans, tmp)
    }
    return

}

php 解法, 执行用时: 8 ms, 内存消耗: 15.7 MB, 提交时间: 2021-07-23 10:08:42

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     public $val = null;
 *     public $left = null;
 *     public $right = null;
 *     function __construct($val = 0, $left = null, $right = null) {
 *         $this->val = $val;
 *         $this->left = $left;
 *         $this->right = $right;
 *     }
 * }
 */
class Solution {

    /**
     * @param TreeNode $root
     * @return Integer[][]
     */
    function levelOrder($root) {
        if ( $root == null ) return [];
        $res = [];
        $queue = [$root];

        while ( !empty($queue) ) {
            $tmp = [];
            $cnt = count($queue);
            for ( $i = 0; $i < $cnt; $i++ ) {
                $node = array_shift($queue);
                $tmp[] = $node->val;
                if ( $node->left ) $queue[] = $node->left;
                if ( $node->right ) $queue[] = $node->right;
            }
            $res[] = $tmp;
        }
        return $res;  // 元素个数即为二叉树的深度
    }
}

python3 解法, 执行用时: 72 ms, 内存消耗: N/A, 提交时间: 2018-08-27 16:26:11

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        vec = []
        if root == None:
            return vec
        from queue import Queue
        q = Queue()
        q.put(root)
        while q.empty() == False:
            s = []
            for i in range(0, q.qsize()):
                temp = q.get()
                if temp.left:
                    q.put(temp.left)
                if temp.right:
                    q.put(temp.right)
                s.append(temp.val)
            vec.append(s)
        return vec