107. Binary Tree Level Order Traversal II 解题记录

题目描述：

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]
]

解题思路：

这题我的想法是遍历每一个节点，把每个节点的数都加到相应层数的一维数组，最后再将二维数组反转。

代码：

/**
 * 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:
    void createArray(TreeNode* cur,int depth,vector<vector<int>>& array){
    //3个形参，cur就不用说了，depth用来计算层数，array用来引用result
        if(!cur)
            return;
        if(depth==array.size()){
        //新的一层就加一个一维数组
            vector<int> temp;
            array.push_back(temp);
        }
        array[depth].push_back(cur->val);//添加
        createArray(cur->left,depth+1,array);
        createArray(cur->right,depth+1,array);
    }
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        createArray(root,0,result);
        return vector<vector<int>>(result.rbegin(), result.rend());//反转结果
    }
};