Question

102. 二叉树的层序遍历

English Version

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。

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

// 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::rc::Rc;
use std::cell::RefCell;
use std::collections::{ VecDeque };
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
  }
}

/**
 * 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;
};