Question

面试题 32 - III. 从上到下打印二叉树 III

题目描述

请实现一个函数按照之字形顺序打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右到左的顺序打印，第三行再按照从左到右的顺序打印，其他行以此类推。

 

例如:
给定二叉树: [3,9,20,null,null,15,7],

  3
   / \
  9  20
  /  \
   15   7

返回其层次遍历结果：

[
  [3],
  [20,9],
  [15,7]
]

 

提示：

1. 节点总数 <= 1000

解法

方法一：BFS

为了实现锯齿形层序遍历，我们每次将当前层的节点添加到结果数组之前，先判断一下当前结果数组的长度，如果是奇数，就将当前层的节点反转一下。之后把当前层的节点添加到结果数组中即可。

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[List[int]]:
    ans = []
    if root is None:
      return ans
    q = deque([root])
    ans = []
    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[::-1] if len(ans) & 1 else t)
    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 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);
        }
      }
      if (ans.size() % 2 == 1) {
        Collections.reverse(t);
      }
      ans.add(t);
    }
    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<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);
        }
      }
      if (ans.size() & 1) {
        reverse(t.begin(), t.end());
      }
      ans.emplace_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)
      }
    }
    if len(ans)&1 == 1 {
      for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
        t[i], t[j] = t[j], t[i]
      }
    }
    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 res = [];
  if (root == null) {
    return res;
  }
  let isEven = false;
  const queue = [root];
  while (queue.length !== 0) {
    const n = queue.length;
    const vals = new Array(n);
    for (let i = 0; i < n; i++) {
      const { val, left, right } = queue.shift();
      vals[i] = val;
      left && queue.push(left);
      right && queue.push(right);
    }
    res.push(isEven ? vals.reverse() : vals);
    isEven = !isEven;
  }
  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<Vec<i32>> {
    let mut res = Vec::new();
    if root.is_none() {
      return res;
    }
    let mut queue = VecDeque::new();
    queue.push_back(root);
    let mut is_even = false;
    while !queue.is_empty() {
      let n = queue.len();
      let mut vals = Vec::with_capacity(n);
      for _ in 0..n {
        let mut node = queue.pop_front().unwrap();
        let mut node = node.as_mut().unwrap().borrow_mut();
        vals.push(node.val);
        if node.left.is_some() {
          queue.push_back(node.left.take());
        }
        if node.right.is_some() {
          queue.push_back(node.right.take());
        }
      }
      if is_even {
        vals.reverse();
      }
      res.push(vals);
      is_even = !is_even;
    }
    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) {
  const ans = [];
  if (!root) {
    return ans;
  }
  const q = [root];
  while (q.length) {
    const t = [];
    for (let n = q.length; n; --n) {
      const { val, left, right } = q.shift();
      t.push(val);
      left && q.push(left);
      right && q.push(right);
    }
    if (ans.length & 1) {
      t.reverse();
    }
    ans.push(t);
  }
  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 IList<IList<int>> LevelOrder(TreeNode root) {
    var ans = new List<IList<int>>();
    if (root == null) {
      return ans;
    }
    var q = new Queue<TreeNode>();
    q.Enqueue(root);
    while (q.Count > 0) {
      var t = new List<int>();
      for (int n = q.Count; n > 0; --n) {
        var node = q.Dequeue();
        t.Add(node.val);
        if (node.left != null) {
          q.Enqueue(node.left);
        }
        if (node.right != null) {
          q.Enqueue(node.right);
        }
      }
      if (ans.Count % 2 == 1) {
        t.Reverse();
      }
      ans.Add(t);
    }
    return ans;
  }
}