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);    }};