LeetCode - Binary Tree Level Order Traversal

Binary Tree Level Order Traversal

2014.1.1 01:53

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

Solution:

　　Level-order traversal of a binary tree is usually performed with BFS and queue. But the return result of this problem is a bit special, we can do it with pre-order traversal as well. The difference is that you calculate the result data in a different order, but same output at last.

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

Accepted code:

// 3CE, 1AC, you fool!!! The code must be bug-free, well.. at least it should compile and runnnnnnnnnnn!
/**
 * 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> > levelOrder(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        // I could use pre-order traversal to do this.
        // Level-order traversal makes sense too.
        
        // 1CE here, declaration of int i is MISSSING
        for(int i = 0; i < result.size(); ++i){
            result[i].clear();
        }
        result.clear();
        
        if(root == nullptr){
            return result;
        }
        
        preOrder(root, 0);
        return result;
    }
private:
    vector<vector<int>> result;
    
    void preOrder(TreeNode *root, int height) {
        if(root == nullptr){
            return;
        }
        // 1CE here,  ) MISSING!!!
        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){
            // 1CE here, spelling error, pre-order preOrder
            // Don't rely on auto-complete, you're spoiled!!!
            preOrder(root->right, height + 1);
        }
    }
};