给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。
示例：
二叉树：[3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回其层次遍历结果：
[
[3],
[9,20],
[15,7]
]最后更新于
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/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function (root) {
if (!root) return [];
const items = []; // 存放所有节点
const queue = [root, null]; // null 简化操作
let levelNodes = []; // 存放每一层的节点
while (queue.length > 0) {
const t = queue.shift();
if (t) {
levelNodes.push(t.val);
if (t.left) {
queue.push(t.left);
}
if (t.right) {
queue.push(t.right);
}
} else {
// 一层已经遍历完了
items.push(levelNodes);
levelNodes = [];
if (queue.length > 0) {
queue.push(null);
}
}
}
return items;
};/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
// 迭代
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
auto ret = vector<vector<int>>();
if (root == nullptr) return ret;
auto q = vector<TreeNode*>();
q.push_back(root);
auto level = 0;
while (!q.empty())
{
auto sz = q.size();
ret.push_back(vector<int>());
for (auto i = 0; i < sz; ++i)
{
auto t = q.front();
q.erase(q.begin());
ret[level].push_back(t->val);
if (t->left != nullptr) q.push_back(t->left);
if (t->right != nullptr) q.push_back(t->right);
}
++level;
}
return ret;
}
};
// 递归
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> v;
levelOrder(root, 0, v);
return v;
}
private:
void levelOrder(TreeNode* root, int level, vector<vector<int>>& v) {
if (root == NULL) return;
if (v.size() < level + 1) v.resize(level + 1);
v[level].push_back(root->val);
levelOrder(root->left, level + 1, v);
levelOrder(root->right, level + 1, v);
}
};# 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: TreeNode) -> List[List[int]]:
"""递归法"""
if root is None:
return []
result = []
def add_to_result(level, node):
"""递归函数
:param level int 当前在二叉树的层次
:param node TreeNode 当前节点
"""
if level > len(result) - 1:
result.append([])
result[level].append(node.val)
if node.left:
add_to_result(level+1, node.left)
if node.right:
add_to_result(level+1, node.right)
add_to_result(0, root)
return result