107. Binary Tree Level Order Traversal II (easy)

Q: 

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

A:

----------------dfs ------------recursive way -----------------

/**
 * 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:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        int d = getDepth(root);
        for(int i =0;i<d; i++)
        {
            vector<int> tmp;
            res.push_back(tmp);
        }
        helper(root, res, d-1);
        return res;
    }
    
    
private:
    int getDepth(TreeNode* const root)
    {
        if(!root)
            return 0;
        return 1 + max(getDepth(root->left), getDepth(root->right));
    }
    
    void helper(TreeNode* const root, vector<vector<int>> &res, int layer)
    {
        if(!root)
            return ;
        helper(root->left, res, layer-1);
        helper(root->right, res, layer-1);
        res[layer].push_back(root->val);        
    }
};

Mistakes:

1:  忘了是bottom-up，因此需要height - 1 - level



------------------BFS   ------记住当前一层的TreeNode---------iterative way -----------
 除了方法名不一样以外，  只在shrink2int方法的最后， 把newAll.add(newAl) 改成newAll.add(0, newAl)----- 即可-------

public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> all = new LinkedList<List<Integer>>();
        if(root == null)
            return all;
        List<TreeNode> curList = new LinkedList<TreeNode>();
        curList.add(root);
        while( !curList.isEmpty()){
            List<TreeNode> newList = new LinkedList<TreeNode>();
            List<Integer> thisLevel = new LinkedList<Integer>();
            while(curList.isEmpty()==false){
                TreeNode cur = curList.remove(0);
                thisLevel.add(cur.val);
                if(cur.left!=null)
                    newList.add(cur.left);
                if(cur.right!=null)
                    newList.add(cur.right);
            }
            all.add(0,thisLevel);
            curList = newList;
        }
        return all;
    }
}