2415

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.

 

Constraints:

• The number of nodes in the tree is in the range [1, 214].
• 0 <= Node.val <= 105
• root is a perfect binary tree.

Related Topics:
Tree, Depth-First Search, Breadth-First Search, Binary Tree

Similar Questions:

• Invert Binary Tree (Easy)

Solution 1. Level-order Traversal

// OJ: https://leetcode.com/problems/reverse-odd-levels-of-binary-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    TreeNode* reverseOddLevels(TreeNode* root) {
        if (!root || !root->left) return root;
        queue<TreeNode*> q{{root}};
        int lv = 0;
        while (q.size()) {
            int cnt = q.size();
            vector<TreeNode*> v;
            while (cnt--) {
                auto n = q.front(); q.pop();
                if (lv % 2) v.push_back(n);
                if (n->left) {
                    q.push(n->left);
                    q.push(n->right);
                }
            }
            if (lv % 2) {
                for (int i = 0, j = v.size() - 1; i < j; ) {
                    swap(v[i++]->val, v[j--]->val);
                }
            }
            ++lv;
        }
        return root;
    }
};

Solution 2. DFS

// OJ: https://leetcode.com/problems/reverse-odd-levels-of-binary-tree
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
    void dfs(TreeNode* a, TreeNode *b, int lv) {
        if (!a) return;
        if (lv % 2) swap(a->val, b->val);
        dfs(a->left, b->right, lv + 1);
        dfs(a->right, b->left, lv + 1);
    }
public:
    TreeNode* reverseOddLevels(TreeNode* root) {
        dfs(root->left, root->right, 1);
        return root;
    }
};