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102. 二叉树的层序遍历

题目描述

给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。

 

示例 1：

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

示例 2：

输入：root = [1]
输出：[[1]]

示例 3：

输入：root = []
输出：[]

 

提示：

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

解法

方法一：BFS

我们可以使用 BFS 的方法来解决这道题。首先将根节点入队，然后不断地进行以下操作，直到队列为空：

  • 遍历当前队列中的所有节点，将它们的值存储到一个临时数组 \(t\) 中，然后将它们的孩子节点入队。
  • 将临时数组 \(t\) 存储到答案数组中。

最后返回答案数组即可。

时间复杂度 \(O(n)\)，空间复杂度 \(O(n)\)。其中 \(n\) 是二叉树的节点个数。

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# 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: Optional[TreeNode]) -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                node = q.popleft()
                t.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t)
        return ans

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int n = q.size(); n > 0; --n) {
                TreeNode node = q.poll();
                t.add(node.val);
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            ans.add(t);
        }
        return ans;
    }
}

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/**
 * Definition for a binary tree node.
 * 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) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        if (!root) return ans;
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            vector<int> t;
            for (int n = q.size(); n; --n) {
                auto node = q.front();
                q.pop();
                t.push_back(node->val);
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
            ans.push_back(t);
        }
        return ans;
    }
};

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/**
 * 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
    }
    q := []*TreeNode{root}
    for len(q) > 0 {
        t := []int{}
        for n := len(q); n > 0; n-- {
            node := q[0]
            q = q[1:]
            t = append(t, node.Val)
            if node.Left != nil {
                q = append(q, node.Left)
            }
            if node.Right != nil {
                q = append(q, node.Right)
            }
        }
        ans = append(ans, t)
    }
    return
}

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/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function levelOrder(root: TreeNode | null): number[][] {
    const ans: number[][] = [];
    if (!root) {
        return ans;
    }
    const q: TreeNode[] = [root];
    while (q.length) {
        const t: number[] = [];
        const qq: TreeNode[] = [];
        for (const { val, left, right } of q) {
            t.push(val);
            left && qq.push(left);
            right && qq.push(right);
        }
        ans.push(t);
        q.splice(0, q.length, ...qq);
    }
    return ans;
}

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// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
        let mut ans = Vec::new();
        if let Some(root_node) = root {
            let mut q = VecDeque::new();
            q.push_back(root_node);
            while !q.is_empty() {
                let mut t = Vec::new();
                for _ in 0..q.len() {
                    if let Some(node) = q.pop_front() {
                        let node_ref = node.borrow();
                        t.push(node_ref.val);
                        if let Some(ref left) = node_ref.left {
                            q.push_back(Rc::clone(left));
                        }
                        if let Some(ref right) = node_ref.right {
                            q.push_back(Rc::clone(right));
                        }
                    }
                }
                ans.push(t);
            }
        }
        ans
    }
}

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/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function (root) {
    const ans = [];
    if (!root) {
        return ans;
    }
    const q = [root];
    while (q.length) {
        const t = [];
        const qq = [];
        for (const { val, left, right } of q) {
            t.push(val);
            left && qq.push(left);
            right && qq.push(right);
        }
        ans.push(t);
        q.splice(0, q.length, ...qq);
    }
    return ans;
};

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