leetcode_102_Binary Tree Level Order Traversal

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102 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]
]

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; *///方法一：BFSstruct Node{  TreeNode *node;  int level;  Node(TreeNode *n , int l) : node(n) , level(l) {}};class Solution {vector<vector<int>> ans;public:    vector<vector<int> > levelOrder(TreeNode *root) {        ans.clear();        if(root == NULL)            return ans;                vector<int> temp;        int Curlevel = -1;                queue<Node> q;        q.push( Node(root , 1) );                while(!q.empty())        {            Node nodeQ = q.front();            q.pop();                        if(nodeQ.node->left)                q.push( Node(nodeQ.node->left , nodeQ.level+1) );            if(nodeQ.node->right)                q.push( Node(nodeQ.node->right , nodeQ.level+1) );                        if(Curlevel != nodeQ.level)            {                if(Curlevel != -1)                    ans.push_back(temp);                                temp.clear();                temp.push_back(nodeQ.node->val);                Curlevel = nodeQ.level;            }            else                temp.push_back(nodeQ.node->val);        }        ans.push_back(temp);                return ans;    }};

//方法二：DFSclass Solution {vector<vector<int>> ans;public:    void levelOrderUtil (TreeNode *root , int depth)    {        if(root==NULL)            return;        if(ans.size() > depth)            ans[depth].push_back(root->val);        else        {            vector<int> a;            a.push_back(root->val);            ans.push_back(a);        }        levelOrderUtil( root->left , depth+1 );        levelOrderUtil( root->right , depth+1 );    }    vector<vector<int> > levelOrder(TreeNode *root) {        ans.clear();        levelOrderUtil( root , 0);        return ans;    }};