LeetCode

Binary Tree Zigzag Level Order Traversal

2014.1.1 02:13

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,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

return its zigzag level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Solution:

　　This problem is a variation from level-order traversal of binary tree, please see the reference here.

　　My solution is to get the level-order traversal first and reverse every other row to get the zig-zag version.

　　Time complexity is O(n), space complexity is O(1), where n is the number of nodes in the tree.

Accepted code:

// 1AC, excellent!!
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode *root) {
        int i;
        
        for(i = 0; i < result.size(); ++i){
            result[i].clear();
        }
        result.clear();
        
        if(root == nullptr){
            return result;
        }
        
        preOrder(root, 0);
        
        for(i = 0; i < result.size(); ++i){
            if(i % 2){
                reverse(result[i], 0, result[i].size() - 1);
            }
        }
        
        return result;
    }
private:
    vector<vector<int>> result;
    
    void preOrder(TreeNode *root, int height) {
        if(root == nullptr){
            return;
        }
        while(result.size() <= height){
            result.push_back(vector<int>());
        }
        result[height].push_back(root->val);
        
        if(root->left != nullptr){
            preOrder(root->left, height + 1);
        }
        if(root->right != nullptr){
            preOrder(root->right, height + 1);
        }
    }
    
    void reverse(vector<int> &num, int left, int right) {
        int i;
        
        if(left >= right){
            return;
        }
        
        int tmp;
        i = left;
        while(i < left + right - i){
            tmp = num[i];
            num[i] = num[left + right - i];
            num[left + right - i] = tmp;
            ++i;
        }
    }
};