【Lintcode】070.Binary Tree Level Order Traversal II

题目：

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).

Example

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

return its bottom-up level order traversal as:

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

题解：

　　迭代法，利用队列，不能用栈。

Solution 1 ()

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode *root) {
        vector<vector<int>> res;
        if (root == nullptr) {
            return res;
        }
        vector<int> v;
        queue<TreeNode*> q;
        q.push(root);
        q.push(nullptr);
        
        while (!q.empty()) {
            TreeNode* cur = q.front();
            q.pop();
            if(cur == nullptr) {
                res.insert(res.begin(),v);
                v.clear();
                if(!q.empty()) {
                    q.push(nullptr);
                }
                continue;
            }
            v.push_back(cur->val);
            if(cur->left != nullptr) {
                q.push(cur->left);
            }
            if(cur->right != nullptr) {
                q.push(cur->right);
            }
        }
        
        return res;
    }
};

Solution 1.1 ()

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode *root) {
        vector<vector<int>> res;
        if (root == nullptr) {
            return res;
        }
        queue<TreeNode*> q;
        q.push(root);
        
        while (!q.empty()) {
            int size = q.size();
            vector<int> v;
            for (int i = 0; i < size; ++i) {
                TreeNode* cur = q.front();
                q.pop();
                if(cur->left != nullptr) {
                    q.push(cur->left);
                }
                if(cur->right != nullptr) {
                    q.push(cur->right);
                }
                v.push_back(cur->val);
            }
            res.insert(res.begin(),v);
        }
        return res;
        
    }
};

　　递归

Solution 2 ()