102. Binary Tree Level Order Traversal

題目

 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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  /  2   3
    /
   4
         5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}". 
 

解析

// 102. Binary Tree Level Order Traversal
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {

        vector<vector<int>> vecs;

        if (!root)
        {
            return vecs;
        }
        queue<TreeNode*> que;
        que.push(root);

        while
 
 (!que.empty())
        {
            int size = que.size();
            vector<int> vec;

            while (size--)
            {
                TreeNode* temp = que.front();
                que.pop();

                vec.push_back(temp->val);
                if (temp->left)
                {
                    que.push(temp->left);
                }
                if
 
 (temp->right)
                {
                    que.push(temp->right);
                }
            }
            vecs.push_back(vec);
        }

        return vecs;
    }

    vector<vector<int>> levelOrder(TreeNode* root) {
        if (!root) { return{}; }
        vector<int> row;
        vector<vector<int> > result;
        queue<TreeNode*> q;
        q.push(root);
        int count = 1;

        while (!q.empty()) {
            if (q.front()->left) { q.push(q.front()->left); }
            if (q.front()->right) { q.push(q.front()->right); }
            row.push_back(q.front()->val), q.pop();
            if (--count == 0) {
                result.emplace_back(row), row.clear();
                count = q.size();
            }
        }
        return result;
    }

    // 遞歸實現;up->bottom
        //類似的：bottom->up: 107. Binary Tree Level Order Traversal II: http://www.cnblogs.com/ranjiewen/p/8253217.html

    vector<vector<int>> ret;
    void buildVector(TreeNode *root, int depth)
    {
        if (root == NULL) return;
        if (ret.size() == depth)
            ret.push_back(vector<int>());

        ret[depth].push_back(root->val);
        buildVector(root->left, depth + 1);
        buildVector(root->right, depth + 1);
    }

    vector<vector<int> > levelOrder(TreeNode *root) {
        buildVector(root, 0);
        return ret;
    }
};

題目來源

• 102. Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal