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：

• 节点总数 <= 1000

解法

Python3

# 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]]:
        if root is None:
            return []
        q = deque()
        res = []
        q.append(root)
        while q:
            size = len(q)
            t = []
            for _ in range(size):
                node = q.popleft()
                t.append(node.val)
                if node.left is not None:
                    q.append(node.left)
                if node.right is not None:
                    q.append(node.right)
            res.append(t if len(res) & 1 == 0 else t[::-1])
        return res

Java

/**
 * 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) {
        if (root == null) return Collections.emptyList();
        Deque<TreeNode> q = new ArrayDeque<>();
        List<List<Integer>> res = new ArrayList<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> t = new ArrayList<>();
            while (size-- > 0) {
                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 ((res.size() & 1) == 1) Collections.reverse(t);
            res.add(t);
        }
        return res;
    }
}

JavaScript

/**
 * 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 [];
  let queue = [root];
  let res = [];
  let depth = 0;
  let dir = true;
  while (queue.length) {
    let len = queue.length;
    for (let i = 0; i < len; i++) {
      let node = queue.shift();
      if (!node) continue;
      if (!res[depth]) res[depth] = [];
      if (dir) {
        res[depth].push(node.val);
      } else {
        res[depth].unshift(node.val);
      }
      queue.push(node.left, node.right);
    }
    depth++;
    dir = !dir;
  }
  return res;
};

Go

func levelOrder(root *TreeNode) [][]int {
    if root == nil {
        return nil
    }
    res := [][]int{}
    queue := []*TreeNode{}
    queue = append(queue,root)
    level := 0
    for len(queue) != 0 {
        size := len(queue)
        ans := []int{}
        //size记录每层大小,level记录层数
        for size > 0 {
            cur := queue[0]
            if level & 1 == 0 {
                ans = append(ans, cur.Val)
            } else {
                ans = append([]int{cur.Val},ans...)
            }

            queue = queue[1:]
            size--
            if cur.Left != nil {
                queue = append(queue, cur.Left)
            }
            if cur.Right != nil {
                queue = append(queue, cur.Right)
            }
        }
        level++
        res = append(res, ans)
    }
    return res
}

C++

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        if (root == NULL) return ans;
        queue<TreeNode*> q;
        q.push(root);
        bool flag = false;
        while (!q.empty()) {
            int n = q.size();
            vector<int> v;
            for (int i = 0; i < n; ++i) {
                TreeNode* node = q.front();
                q.pop();
                v.emplace_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            if (flag) reverse(v.begin(), v.end());
            flag = !flag;
            ans.emplace_back(v);
        }
        return ans;
    }
};

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