L 366. Find Leaves of Binary Tree --------M

Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

 

Example:

Input: [1,2,3,4,5]
  
          1
         / \
        2   3
       / \     
      4   5    

Output: [[4,5,3],[2],[1]]

 

Explanation:

1. Removing the leaves [4,5,3] would result in this tree:

          1
         / 
        2          

 

2. Now removing the leaf [2] would result in this tree:

          1          

 

3. Now removing the leaf [1] would result in the empty tree:

          []         

[[3,5,4],[2],[1]], [[3,4,5],[2],[1]], etc, are also consider correct answers since per each level it doesn't matter the order on which elements are returned.

A:

就是简单的递归，然后返回 到 leaf 的距离

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> res;
        helper(root, res);
        return res;
    }
private:
    int helper(TreeNode* root, vector<vector<int>>& res ){ // return depth to lowest leaf
        if(not root)
            return -1;
        int l = helper(root->left, res);
        int r = helper(root->right, res);
        int level = max(l, r)+1;
        if(res.size() < level + 1){
            res.push_back(vector<int>());
        }
        res[level].push_back(root->val);
        return level;
    }   
};

Mistakes：

level 没有+1