本问题对应的 leetcode 原文链接：剑指 Offer 32 – I. 从上到下打印二叉树
问题描述
从上到下打印出二叉树的每个节点，同一层的节点按照从左到右的顺序打印。
例如:
给定二叉树: [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回：
[3,9,20,15,7]
提示：
节点总数 <= 1000
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public int[] levelOrder(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)
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
from collections import deque
class Solution(object):
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if not root:
return []
queue = deque([root])
res = []
while queue:
node = queue.popleft()
res.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return res
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> levelOrder(TreeNode* root) {
vector<int> res;
if (!root) {
return res;
}
queue<TreeNode*> q{{root}};
while (!q.empty()) {
auto t = q.front();
q.pop();
res.push_back(t->val);
if (t->left) {
q.push(t->left);
}
if (t->right) {
q.push(t->right);
}
}
return res;
}
};
Go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func levelOrder(root *TreeNode) []int {
if root == nil {
return []int{}
}
var res []int
queue := []*TreeNode{root}
for len(queue) > 0 {
node := queue[0]
queue = queue[1:]
res = append(res, node.Val)
if node.Left != nil {
queue = append(queue, node.Left)
}
if node.Right != nil {
queue = append(queue, node.Right)
}
}
return res
}
JS
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var levelOrder = function(root) {
if (root == null) {
return [];
}
var queue = [];
var res = [];
queue.push(root);
while (queue.length > 0) {
var t = queue.shift();
res.push(t.val);
if (t.left != null) queue.push(t.left);
if (t.right != null) queue.push(t.right);
}
return res;
};
shuaidi