Binary Tree Level Order Traversal II -- leetcode
2015-05-08 16:16
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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
return its bottom-up level order traversal as:
基本思路:
同问题 Binary Tree Level
Order Traversal 一样,
在结果返回前,作一个reverse操作。
For example:
Given binary tree
{3,9,20,#,#,15,7},
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
基本思路:
同问题 Binary Tree Level
Order Traversal 一样,
在结果返回前,作一个reverse操作。
/** * 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> > ans; dfs(ans, root, 0); reverse(ans.begin(), ans.end()); return ans; } void dfs(vector<vector<int> > &ans, TreeNode *root, int level) { if (!root) return; if (level == ans.size()) ans.push_back(vector<int>()); ans[level].push_back(root->val); dfs(ans, root->left, level+1); dfs(ans, root->right, level+1); } };
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