Binary_Tree_Level_Order_Traversal.txt

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

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > lists;
        queue<TreeNode *> current;
        if (root!=NULL) current.push(root);
        while(!current.empty()){
            vector<int> list;//When we do this, we create a BRAND NEW list.
            int size = current.size();
            for (int i=0; i<size; ++i){
                TreeNode * t = current.front();
                list.push_back(t->val);
                current.pop();
                if (t->left!=NULL)
                current.push(t->left);
                if (t->right!=NULL)
                current.push(t->right);
            }
            lists.push_back(list);
        }
        return lists;
    }
};

REMARK:
1.Hard question. IDEA: We go through level by level using a while loop. "current" stores and only stores all the node in one level. Then we store the value in "current" to the "list" by using "size = current.empty" to indicate how many nodes there are in this level and a for loop.
2.Time: O(n).
3. For vector,we use push_back(). For queue, we use push(),pop(),front(),empty().