LeetCode--Binary Tree Level Order Traversal II

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：和之前Binary Tree Level Order Traversal这个一样，只不过最后reverse反转一下，倒序。

方法一：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) {
        vector<vector<int>>result;
        traverse(root,result,1);
        reverse(result.begin(),result.end());
        return result;
    }
    void traverse(TreeNode* root,vector<vector<int>>&result,int level){
        if(!root) return;
        if(level>result.size()){
            result.push_back(vector<int>());
        }
        result[level-1].push_back(root->val);
        traverse(root->left,result,level+1);
        traverse(root->right,result,level+1);
    }
};