Leetcode题解（1）：L102/Binary Tree Level Order Traversal

L102: Binary Tree Level Order Traversal

　　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]

]

解题思路:与BFS差不多，只不过要增加一个挡板标识区分每一层

//Definition for a binary tree node.
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) {
        vector<vector<int>> levelOrdVec;

        if(root == 0)
            return levelOrdVec;

        queue<TreeNode*> nodeQueue;
        TreeNode* mark = nullptr;

        nodeQueue.push(root);

        while(!nodeQueue.empty())
        {
            nodeQueue.push(mark);
            vector<int> ordVec;
            TreeNode* node = nullptr;
            while((node = nodeQueue.front()) != mark)
            {
                nodeQueue.pop();
                ordVec.push_back(node->val);
                if(node->left) nodeQueue.push(node->left);
                if(node->right) nodeQueue.push(node->right);
            }
            nodeQueue.pop(); //pop mark

            levelOrdVec.push_back(ordVec);
        }
        return move(levelOrdVec);
    }
};