leetcode笔记：Binary Tree Level Order Traversal II

一. 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,#,#,15,7},

   3
  /   9 20
    /    15  7

二. 题目分析

由于使用了vector，这一题只需Binary Tree Level Order Traversal的基础上加一句reverse(result.begin(), result.end()) 即可。印象中编程之美上有这题。

三. 示例代码

#include <iostream>
#include <vector>

using namespace std;

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;
        orderTraversal(root, 1, result);
        reverse(result.begin(), result.end()); // 新增
        return result;
    }

private:
    void orderTraversal(TreeNode *root, size_t level, vector<vector<int> > & result)
    {
        if (root == NULL) return;

        if (level > result.size())
            result.push_back(vector<int>());

        result[level - 1].push_back(root->val);

        orderTraversal(root->left, level + 1, result);
        orderTraversal(root->right, level + 1, result);
    }
};

四. 小结

解法同Binary Tree Level Order Traversal，同样有多种解法，还需认真研究。