102.binary-tree-level-order-traversal

Statement

• Difficulty: Medium
• Tag: `树` `广度优先搜索` `二叉树`

``````输入：root = [3,9,20,null,null,15,7]

``````

``````输入：root = [1]

``````

``````输入：root = []

``````

• 树中节点数目在范围 `[0, 2000]`
• `-1000 <= Node.val <= 1000`

Given the `root` of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example 1:

``````Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
``````

Example 2:

``````Input: root = [1]
Output: [[1]]
``````

Example 3:

``````Input: root = []
Output: []
``````

Constraints:

• The number of nodes in the tree is in the range `[0, 2000]`.
• `-1000 <= Node.val <= 1000`

Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

#ifdef LOCAL

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

#endif

class Solution {
public:
vector<vector<int>> res;
void dfs(TreeNode *rt, int dep) {
if (!rt) {
return;
}

if (res.size() < dep + 1) {
res.push_back(vector<int>());
}

res[dep].push_back(rt->val);
dfs(rt->left, dep + 1);
dfs(rt->right, dep + 1);
}

vector<vector<int>> levelOrder(TreeNode *root) {
dfs(root, 0);
return res;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````