*Leetcode 102. Binary Tree Level Order Traversal

https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/

1、BFS解法：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */struct DTreeNode {    TreeNode *node;    int dep;    DTreeNode(int d, TreeNode *r): dep(d), node(r) {}};class Solution {public:    vector<vector<int> > levelOrder(TreeNode* root) {        vector < vector<int> > ret;                if (root == NULL) return ret;                vector < int > dep0;        queue < DTreeNode* > q;        q.push( new DTreeNode(0, root) );                       while (! q.empty()) {            DTreeNode * tp = q.front();            q.pop();            int dep = tp->dep;                        if (dep >= ret.size()) {                ret.push_back( vector<int>{} );            }                        ret[dep].push_back( tp->node->val );                        if (tp->node->left) {                q.push( new DTreeNode(dep+1, tp->node->left) );            }                        if (tp->node->right) {                q.push( new DTreeNode(dep+1, tp->node->right) );            }        }                return ret;    }};

2、先序遍历

这个明显更快，应该是因为没有new变量等过程吧

/** * 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:        void dfs(vector< vector<int> > &ans, int dep, TreeNode *root ) {        if (root == NULL) return;        if (dep == ans.size()) {            ans.push_back( vector<int>{} );        }                ans[dep].push_back(root->val);        dfs(ans, dep+1, root->left);        dfs(ans, dep+1, root->right);    }        vector<vector<int>> levelOrder(TreeNode* root) {        vector< vector<int> > ans;        dfs( ans, 0, root );        return ans;    }};