二叉树之字形水平阶遍历的这种算法的时间复杂度是多少？

Question

二叉树Z字形级别的顺序遍历
给定一个二叉树，返回其节点值的Z字形级别顺序遍历。(即从左到右，然后从右到左进入下一级并在两者之间交替(。

例如：给定二叉树{3,9,20，#，#，15,7}，

        3 
       /  
      9   20 
     /  
   15   7 

返回其Z字形级别的顺序遍历为：

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

---

我个人认为，
时间复杂性=O(n*height(，n是节点数，height是给定二叉树的高度。

   getHeight()             => O(n) 
   traverseSpecificLevel() => O(n)
   reverseVector()         => O(n)
   swap()                  => O(1)

---

C++

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <vector>
using namespace std;
class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int>> list;
        // Input validation.
        if (root == NULL) return list;
        // Get the height of the binary tree.
        int height = getHeight(root);
        bool left_to_right = true;
        for (int level = 0; level <= height; level ++) {
            vector<int> subList;
            traverseSpecificLevel(root, level, subList);
            if (left_to_right == true) {
                // Add subList into list.
                list.push_back(subList);
                // Update left_to_right flag.
                left_to_right = false;
            } else {
                // Reverse subList.
                reverseVector(subList);
                // Add reversed subList into list.
                list.push_back(subList);
                // Update left_to_right flag.
                left_to_right = true;
            }
        }
        return list;
    }
    int getHeight(TreeNode *root) {
        // Base case.
        if (root == NULL || (root->left == NULL && root->right == NULL)) return 0;
        else return 1 + max(getHeight(root->left), getHeight(root->right));
    }
    void traverseSpecificLevel(TreeNode *root, int level, vector<int> &subList) {
        // Base case.
        if (root == NULL) return;
        if (level == 0) {
            subList.push_back(root->val);
            return;
        }
        // Do recursion.
        traverseSpecificLevel(root->left, level - 1, subList);
        traverseSpecificLevel(root->right, level - 1, subList);
    }
    void reverseVector(vector<int> &list) {
        // Input validation.
        if (list.size() <= 1) return;
        int start = 0;
        int end = list.size() - 1;
        while (start < end) {
            swap(list, start, end);
            start ++;
            end --;
        }
    }
    void swap(vector<int> &list, int first, int second) {
        int tmp = list[first];
        list[first] = list[second];
        list[second] = tmp;
    }
};

Answer 1

您可以在线性时间内完成。创建大小为max_height的向量>结果。递归遍历树，保持节点的级别。对于每个节点，将其值推回到结果[级别]。而不仅仅是颠倒结果[1]，结果[3]。

顺便说一下，有一个swap(x,y)函数和reverse(a.begin(), a.end())函数(其中a是一个向量(，您可以使用它们，而不是自己实现它们。包括algorithm。