Binary Tree Level Order Traversal & II

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 a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
   private:
     vector<vector<int> > ret;            //这里是C++，数据写在外面
 public:
     void solve(int dep, TreeNode *root)
     {
         if (root == NULL)
             return;
             
         if (ret.size() > dep)
         {
             ret[dep].push_back(root->val);        //根据每一层来压入数据
         }
         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;
     }
 };

 

 

Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

/**
 * 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 {
    vector<vector<int>> ret, out;
public:
    void getDepVal(int deep, TreeNode* root)
    {
        if(root == NULL)
            return;
        else if(ret.size() > deep)
            ret[deep].push_back(root->val);
        else
            {
                vector<int> tmp;
                tmp.push_back(root->val);
                ret.push_back(tmp);
            }
        getDepVal(deep+1, root->left);
        getDepVal(deep+1, root->right);
    }
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        getDepVal(0, root);
        reverse(ret.begin(), ret.end());
        return ret;
    }
};