LeetCode 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},

DFS遍历方法：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<vector<int> > ret;
public:
    void solve(int dep, TreeNode *root)
    {
        if (root == NULL)
            return;
            
        if (ret.size() > dep)
        {
            ret[dep].push_back(root->val);
        }
        else
        {
            vector<int> a;
            a.push_back(root->val);
            ret.push_back(a);
        }
        
        solve(dep + 1, root->left);
        solve(dep + 1, root->right);
    }
    
    vector<vector<int> > levelOrder(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ret.clear();
        solve(0, root);
        
        return ret;
    }
};

BFS遍历打印

/**
 * Definition for binary tree
 * 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;
    }
};