ama-non

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],

return its bottom-up level order traversal as:

分析：

在Level Order Traversal的基础上reverse()一下即可。

思路一：递归

class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int>> result;
        traverse(root, 1, result);
        std::reverse(result.begin(), result.end());
        // 比Level Order Traversal多此一行
        return result;
    }

    void traverse(TreeNode *root, size_t level, vector<vector<int>> &result) {
        if (!root) return;

        if (level > result.size())
            result.push_back(vector<int>());

        result[level-1].push_back(root->val);
        traverse(root->left, level+1, result);
        traverse(root->right, level+1, result);
    }
};

时间复杂度o(n),空间复杂度o(n)

思路二：迭代

class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > result;
        if(root == nullptr) return result;

        queue<TreeNode*> current, next;
        vector<int> level; // elments in level level

        current.push(root);
        while (!current.empty()) {
            while (!current.empty()) {
                TreeNode* node = current.front();
                current.pop();
                level.push_back(node->val);
                if (node->left != nullptr) next.push(node->left);
                if (node->right != nullptr) next.push(node->right);
            }
            result.push_back(level);
            level.clear();
            swap(next, current);
        }
        reverse(result.begin(), result.end()); // 比上一题多此一行
        return result;
    }
};

时间复杂度o(n),空间复杂度o(1)

参考来源：

• https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/binary-tree/traversal/binary-tree-level-order-traversal-ii.html