【LeetCode】2415. Reverse Odd Levels of Binary Tree 解題報告

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2415. Reverse Odd Levels of Binary Tree / Medium

Given the root of a perfect binary tree, reverse the node values at each odd level of the tree.

  • For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18], then it should become [18,29,11,7,4,3,1,2].
    Return the root of the reversed tree.

A binary tree is perfect if all parent nodes have two children and all leaves are on the same level.

The level of a node is the number of edges along the path between it and the root node.

Example 1:

Input: root = [2,3,5,8,13,21,34]
Output: [2,5,3,8,13,21,34]
Explanation:
The tree has only one odd level.
The nodes at level 1 are 3, 5 respectively, which are reversed and become 5, 3.

Example 2:

Input: root = [7,13,11]
Output: [7,11,13]
Explanation:
The nodes at level 1 are 13, 11, which are reversed and become 11, 13.

Example 3:

Input: root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
Output: [0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
Explanation:
The odd levels have non-zero values.
The nodes at level 1 were 1, 2, and are 2, 1 after the reversal.
The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1 after the reversal.

Comstraints

  • The number of nodes in the tree is in the range [1, 2^14].
  • 0 <= Node.val <= 10^5
  • root is a perfect binary tree.

Solution 1: BFS

思路

利用 BFS 可以很直覺的拿到每一層的 node，再將其 val swap 就好。
相關題目
102. Binary Tree Level Order Traversal

103. Binary Tree Zigzag Level Order Traversal

效能

Complexity

  • Time Complexity: O(N), where N is the number of nodes
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    TreeNode* reverseOddLevels(TreeNode* root) {
        int level = 1;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            int size = q.size();
            vector<TreeNode*> nodes;
            while(size--) {
                nodes.emplace_back(q.front()); q.pop();
                if(nodes.back()->left) q.push(nodes.back()->left);
                if(nodes.back()->right) q.push(nodes.back()->right);
            }
            
            if(level % 2 == 0) {
                int l = 0, r = nodes.size()-1;
                while(l < r) {
                    swap(nodes[l]->val, nodes[r]->val);
                    ++l; --r;
                }
            }
            ++level;
        }
        return root;
    }
};

Solution 2: DFS

思路

利用 DFS 搭配 level 就簡單很多了，要注意的是 reverse 的 swap 應該是從外而內
所以是

reverse(leftN->left, rightN->right, level+1);
reverse(leftN->right, rightN->left, level+1);

效能

Complexity

  • Time Complexity: O(N), where N is the number of nodes
  • Space Complexity: O(N) or O(H), where H is the height of the given tree

LeetCode Result

Code

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class Solution {
public:
    TreeNode* reverseOddLevels(TreeNode* root) {
        reverse(root->left, root->right, 2);
        return root;
    }
    
    void reverse(TreeNode* leftN, TreeNode* rightN, int level) {
        if(!leftN && !rightN) return;
        if(level % 2 == 0) swap(leftN->val, rightN->val);
        reverse(leftN->left, rightN->right, level+1);
        reverse(leftN->right, rightN->left, level+1);
    }
};