【劍指Offer系列32-3】從上到下打印二叉樹3

題目

請實現一個函數按照之字形順序打印二叉樹，即第一行按照從左到右的順序打印，第二層按照從右到左的順序打印，第三行再按照從左到右的順序打印，其他行以此類推。

例如:
給定二叉樹: [3,9,20,null,null,15,7],

3
/
9 20
/
15 7

返回其層次遍歷結果：

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

提示：
節點總數 <= 1000

代碼

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

# 思路：
# BFS+deque，奇偶層分別入隊尾，隊頭
# 複雜度：
# O(N)
class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root: return []
        res, Q = [], collections.deque()
        Q.append(root) # 根入隊
        while Q:
            tmp = [] # 暫存列表
            # 打印奇數層
            for _ in range(len(Q)):
                # 從左向右打印
                node = Q.popleft()
                tmp.append(node.val)
                # 先左子樹後右子樹
                if node.left: Q.append(node.left)
                if node.right: Q.append(node.right)
            res.append(tmp)
            
            if not Q: break # 若爲空則提前跳出
                
            # 打印偶數層
            tmp = []
            for _ in range(len(Q)):
                # 從右向左打印
                node = Q.pop()
                tmp.append(node.val)
                # 先右子樹後左子樹
                if node.right: Q.appendleft(node.right)
                if node.left: Q.appendleft(node.left)
            res.append(tmp)
            
        return res

C++

/**
 * 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>> res;
        if(!root) return res;
        deque<TreeNode*> Q;
        Q.push_back(root);
        while(!Q.empty()){
            vector<int> tmp;
            for(int i = Q.size(); i > 0; i--){
                TreeNode* node = Q.front();
                Q.pop_front();
                tmp.push_back(node->val);
                if(node->left) Q.push_back(node->left);
                if(node->right) Q.push_back(node->right);    
            }
            res.push_back(tmp);
            tmp.clear();
            
            if(Q.empty()) break;

            for(int i = Q.size(); i > 0; i--){
                TreeNode* node = Q.back();
                Q.pop_back();
                tmp.push_back(node->val);
                if(node->right) Q.push_front(node->right);
                if(node->left) Q.push_front(node->left);
            }
            res.push_back(tmp);
        }
        return res;
    }
};