【LeetCode笔记】Binary Tree Level Order Traversal II 二叉树按层遍历，反向输出

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

思路：

这题不难，按层遍历一个queue就行，如果要反向输出，再用一个stack调整一下顺序就好了！

/**
 * 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>> v_e;
        if (root==NULL) 
            return v_e;
        else{
            TreeNode* t;
            queue<TreeNode*> q;
            q.push(root);
            t = q.front();
            stack<vector<int>> s;
            while (!q.empty()){
                int size = q.size();
                vector<int> v;
                for (int i = 0; i < size;i++){
                    t = q.front();
                    v.push_back(t->val);
                    q.pop();
                    if(t->left!=NULL) 
                        q.push(t->left);
                    if(t->right!=NULL) 
                        q.push(t->right);
                }
                s.push(v);
            }
            while(!s.empty()){
                v_e.push_back(s.top());
                s.pop();
            }
            return v_e;
        }
    }
};