花花酱 LeetCode 103. Binary Tree Zigzag Level Order Traversal

zxi — Wed, 21 Aug 2019 05:10:56 +0000

Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree [3,9,20,null,null,15,7],

return its zigzag level order traversal as:

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

Solution 1: DFS

in order traversal using DFS and reverse the result of even levels.

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua, 0ms, 15.1 MB
class Solution {
public:
  vector > zigzagLevelOrder(TreeNode *root) {
    vector> ans;
    function dfs = [&](TreeNode* r, int d) {
      if (!r) return;
      while (ans.size() <= d) ans.push_back({});      
      ans[d].push_back(r->val);      
      dfs(r->right, d + 1);
      dfs(r->left, d + 1);
    };
    dfs(root, 0);    
    for (int i = 0; i < ans.size(); i += 2)
      reverse(begin(ans[i]), end(ans[i]));      
    return ans;
  }
};

Solution 2: BFS

Expend/append in order for even levels and doing that in reverse order for odd levels.

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua
class Solution {
public:
  vector> zigzagLevelOrder(TreeNode *root) {
    if (!root) return {};
    vector> ans;
    deque q;
    q.push_back(root);
    int d = 0;
    while (q.size()) {
      ans.push_back({});    
      auto cur = &ans.back();
      deque next;
      while (q.size()) {
        if (d % 2) {
          TreeNode* node = q.back();
          q.pop_back();
          cur->push_back(node->val);
          if (node->right) next.push_front(node->right);
          if (node->left) next.push_front(node->left);          
        } else {
          TreeNode* node = q.front();
          q.pop_front();
          cur->push_back(node->val);
          if (node->left) next.push_back(node->left);
          if (node->right) next.push_back(node->right);          
        }
      }
      ++d;
      q.swap(next);
    }
    return ans;
  }
};

花花酱 LeetCode 6. ZigZag Conversion

zxi — Wed, 12 Sep 2018 15:54:49 +0000

Problem

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:
Input: s = “PAYPALISHIRING”, numRows = 3 Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

Solution: Simulation

Store the zigzag results for each row and then output row by row.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua
class Solution {
public:
  string convert(string s, int nRows) {
    if (nRows == 1) return s;

    vector ss(nRows);
    int l = s.length();
    int x = 0, y = 0, i = 0;
    bool down = true;
    while (i < l) {
      ss[y] += s[i];
      if (down) {
        ++y;
        if (y == nRows) {
            down = false;
            y -=2;
        }
      } else {
        --y;
        if (y < 0) {
            down = true;
            y = 1;
        }
      }
      i++;
    }

    string ans;
    for(const string& r : ss)
      ans += r;
    return ans;
  }
};

The post 花花酱 LeetCode 6. ZigZag Conversion appeared first on Huahua's Tech Road.

zigzag Archives - Huahua's Tech Road