leetcode笔记:Binary Tree Level Order Traversal II
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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。
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