leetcode解题: Binary Tree Level Order Traversal II (107)

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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]
]

解法1：BST + Two container

用按层遍历（one queue）的办法遍历得出每一层的vector，然后放入一个stack，最后按顺序读出vector中的值即可。
C++

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/**
 * 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>> temp;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            int k = q.size();
            vector<int> level;
            for (int i = 0; i < k; ++i) {
                TreeNode* node = q.front();
                q.pop();
                level.push_back(node->val);
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
            temp.push(level);
        }
        while (!temp.empty()) {
            res.push_back(temp.top());
            temp.pop();
        }
        return res;
    }
};

Java

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