花花酱 LeetCode 117. Populating Next Right Pointers in Each Node II

zxi — Mon, 29 Nov 2021 08:14:24 +0000

Given a binary tree

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example 1:

Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 6000].
-100 <= Node.val <= 100

Follow-up:

You may only use constant extra space.
The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.

Solution -2: Group nodes into levels

Use pre-order traversal to group nodes by levels.
Connects nodes in each level.

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

C++

// Author: Huahua
class Solution {
public:
  Node* connect(Node* root) {
    vector> levels;
    dfs(root, 0, levels);
    for (auto& nodes : levels)
      for(int i = 0; i < nodes.size() - 1; ++i)
        nodes[i]->next = nodes[i+1]; 
    return root;
  }
private:
  void dfs(Node *root, int d, vector>& levels) {
    if (!root) return;
    if (levels.size() < d+1) levels.push_back({});
    levels[d].push_back(root);
    dfs(root->left, d + 1, levels);
    dfs(root->right, d + 1, levels);
  }
};

Solution -1: BFS level order traversal

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

C++

// Author: Huahua
class Solution {
public:
  Node* connect(Node* root) {
    if (!root) return root;
    queue q{{root}};
    while (!q.empty()) {
      int size = q.size();
      Node* p = nullptr;
      while (size--) {
        Node* t = q.front(); q.pop();
        if (p) 
          p->next = t, p = p->next;
        else
          p = t;
        if (t->left) q.push(t->left);
        if (t->right) q.push(t->right);
      }
    }
    return root;
  }
};

Solution 1: BFS w/o extra space

Populating the next level while traversing current level.

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua
class Solution {
public:
  Node* connect(Node* root) {
    Node* p = root;        
    while (p) {
      Node dummy;
      Node* c = &dummy;
      while (p) {
        if (p->left)
          c->next = p->left, c = c->next;
        if (p->right)
          c->next = p->right, c = c->next;
        p = p->next;
      }
      p = dummy.next;
    }
    return root;
  }
};

The post 花花酱 LeetCode 117. Populating Next Right Pointers in Each Node II appeared first on Huahua's Tech Road.

花花酱 LeetCode 107. Binary Tree Level Order Traversal II

zxi — Thu, 30 Jan 2020 02:15:09 +0000

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

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

return its bottom-up level order traversal as:

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

Solution: Recursion

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

C++

// Author: Huahua
class Solution {
public:
    vector> levelOrderBottom(TreeNode *root) {
      vector> ans;
      levelOrderBottom(root, 0, ans);
      reverse(ans.begin(), ans.end());
      return ans;
    }
private:
    void levelOrderBottom(TreeNode *root, int depth, vector>& ans) {
      if (!root) return;
      while (ans.size() <= depth) ans.push_back({});
      ans[depth].push_back(root->val);
      levelOrderBottom(root->left, depth + 1, ans);
      levelOrderBottom(root->right, depth + 1, ans);
    }
};

level order Archives - Huahua's Tech Road

花花酱 LeetCode 117. Populating Next Right Pointers in Each Node II

Solution -2: Group nodes into levels

C++

Solution -1: BFS level order traversal

C++

Solution 1: BFS w/o extra space

C++

花花酱 LeetCode 107. Binary Tree Level Order Traversal II

Solution: Recursion

C++

Related Problems