102 - Binary Tree Level Order Traversal(BFS)

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.

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思路分析：

基于TreeNode新构造一个数据结构Node

struct TreeNode
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

struct Node
{
    TreeNode *node;
    int level;
    Node(){}
    Node(TreeNode *n, int l):node(n), level(l){}
};

class Solution {
private:
    vector<vector<int> > ret;
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ret.clear();
        
        if (root == NULL)
            return ret;
        
        queue<Node> q;
        
        q.push(Node(root, 0));
        
        vector<int> a;
        int curLevel = -1;
        
        while(!q.empty())
        {
            Node node = q.front();
            if (node.node->left)
                q.push(Node(node.node->left, node.level + 1));
            if (node.node->right)
                q.push(Node(node.node->right, node.level + 1));
                
            if (curLevel != node.level)
            {
                if (curLevel != -1)
                    ret.push_back(a);
                curLevel = node.level;
                a.clear();
                a.push_back(node.node->val);                
            }
            else
                a.push_back(node.node->val);
                
            q.pop();
        }
        
        ret.push_back(a);
        
        return ret;
    }
};