[Leetcode 77] 102 Binary Tree Level Order Traversal

Problem:

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]
]

Analysis:

The problem is an extension of the original level-order traversal problem. We need an extra sentinal node to signal the end of a level. 

Each time we encounter it, we know that a level is ended so we push the partial_result_vector into the result vector and then immediately push it into the queue again.

The quit condition switch from queue.empty() to there's only the sentinal node in the queue.

Code:

/**
 * 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> > res;
    if (root == NULL)
        return res;

    queue<TreeNode *> q;
    q.push(root);

    TreeNode *tmp, *sentl;

    sentl = new TreeNode((1<<31));
    sentl->left = sentl->right = NULL;
    q.push(sentl);

    vector<int> pres;
    while (!(q.size()==1 && q.front() == sentl)) {
        tmp = q.front();
        q.pop();

        if (tmp == sentl) {
            res.push_back(pres);
            pres.clear();
            q.push(sentl);
        } else {
            pres.push_back(tmp->val);

            if (tmp->left != NULL)
                q.push(tmp->left);

            if (tmp->right != NULL)
                q.push(tmp->right);
        }

    }

    res.push_back(pres);

    return res;
}
};