【LeetCode】107. 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).

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]
]

提示：

此题要求从下往上，从左往右逐行输出二叉树的节点数值，这里提供两种解法，分别是递归和迭代。

代码：

递归：

/**
 * 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) {
        levelOrder(root, 0);
        reverse(res.begin(), res.end());
        return res;
    }
    
private:
    void levelOrder(TreeNode* root, int currLevel) {
        if (root == NULL) return;

        if (res.empty() || currLevel > (res.size() - 1)) res.push_back(vector<int>());

        res[currLevel].push_back(root->val);

        levelOrder(root->left, currLevel + 1);
        levelOrder(root->right, currLevel + 1);
    }
    vector<vector<int>> res;
};

迭代:

/**
 * 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;
        if (!root) return res;
        stack<vector<int>> s;
        vector<int> v;
        TreeNode *node;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                node = q.front(); q.pop();
                v.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            s.emplace(move(v));
            v.clear();
        }
        while(!s.empty()) {
            res.emplace_back(move(s.top()));
            s.pop();
        }
        return res;
    }
};