[LeetCode107 ] Binary Tree Level Order Traversal II 二叉树层序遍历之二

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algorithms	Easy (65.56%)	257	-

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给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

例如：
给定二叉树 [3,9,20,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

返回其自底向上的层次遍历为：

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

方法一：
本题是102 的简单变形，代码如下：

/*
 * @Descripttion: 
 * @version: 
 * @Author: wangxf
 * @Date: 2020-07-07 23:09:12
 * @LastEditors: Do not edit
 * @LastEditTime: 2020-07-07 23:10:30
 */ 
/*
 * @lc app=leetcode.cn id=107 lang=cpp
 *
 * [107] 二叉树的层次遍历 II
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct NodeInfo
{
    TreeNode* ptr = nullptr;
    int level = 0;
    NodeInfo(TreeNode* node_ptr,int node_level):
    ptr(node_ptr),level(node_level)
    {}
};
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root)
    {
        std::vector<vector<int>> res;
        if(!root) return res;
        std::queue<NodeInfo> q;
        NodeInfo root_node(root,0);
        q.push(root_node);
        while(!q.empty())
        {
            NodeInfo cur_node = q.front();
            q.pop();
            int cur_level = cur_node.level;
            int cur_node_val = cur_node.ptr->val;
            if(res.size()<cur_level+1)
            {
                vector<int> tempVec;
                res.push_back(tempVec);
            }
            res[cur_level].push_back(cur_node_val);
            if(cur_node.ptr->left)
            {
               NodeInfo leftNode(cur_node.ptr->left,cur_level+1);
               q.push(leftNode);
            }
            if(cur_node.ptr->right)
            {
               NodeInfo rightNode(cur_node.ptr->right,cur_level+1);
               q.push(rightNode);
            }
        }
        std::reverse(res.begin(),res.end());
        return res;

    }
};
// @lc code=end