Question

面试题 32 - I. 从上到下打印二叉树

题目描述

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：

1. 节点总数 <= 1000

解法

方法一：BFS

我们可以通过 BFS 遍历二叉树，将每一层的节点值存入数组中，最后返回数组即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点数。

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

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

/**
 * 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) {
    if (!root) {
      return {};
    }
    vector<int> ans;
    queue<TreeNode*> q{{root}};
    while (!q.empty()) {
      for (int n = q.size(); n; --n) {
        auto node = q.front();
        q.pop();
        ans.push_back(node->val);
        if (node->left) {
          q.push(node->left);
        }
        if (node->right) {
          q.push(node->right);
        }
      }
    }
    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 {
    for n := len(q); n > 0; n-- {
      node := q[0]
      q = q[1:]
      ans = append(ans, node.Val)
      if node.Left != nil {
        q = append(q, node.Left)
      }
      if node.Right != nil {
        q = append(q, node.Right)
      }
    }
  }
  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 res = [];
  if (root == null) {
    return res;
  }
  const queue = [root];
  while (queue.length !== 0) {
    const { val, left, right } = queue.shift();
    res.push(val);
    left && queue.push(left);
    right && queue.push(right);
  }
  return res;
}

// 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<i32> {
    let mut res = Vec::new();
    let mut queue = VecDeque::new();
    if let Some(node) = root {
      queue.push_back(node);
    }
    while let Some(node) = queue.pop_front() {
      let mut node = node.borrow_mut();
      res.push(node.val);
      if let Some(l) = node.left.take() {
        queue.push_back(l);
      }
      if let Some(r) = node.right.take() {
        queue.push_back(r);
      }
    }
    res
  }
}

/**
 * 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) {
    return [];
  }
  const q = [root];
  const ans = [];
  while (q.length) {
    for (let n = q.length; n; --n) {
      const { val, left, right } = q.shift();
      ans.push(val);
      left && q.push(left);
      right && q.push(right);
    }
  }
  return ans;
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left;
 *   public TreeNode right;
 *   public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
  public int[] LevelOrder(TreeNode root) {
    if (root == null) {
      return new int[]{};
    }
    Queue<TreeNode> q = new Queue<TreeNode>();
    q.Enqueue(root);
    List<int> ans = new List<int>();
    while (q.Count != 0) {
      int x = q.Count;
      for (int i = 0; i < x; i++) {
        TreeNode node = q.Dequeue();
        ans.Add(node.val);
        if (node.left != null) {
          q.Enqueue(node.left);
        }
        if (node.right != null) {
          q.Enqueue(node.right);
        }
      }
    }
    return ans.ToArray();
  }
}