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107. 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] ]
解題思路:
此題與Binary Tree Level Order Traversal相似,只是最后的結(jié)果有一個(gè)反轉(zhuǎn)。
參考 http://qiaopeng688.blog.51cto.com/3572484/1834819
代碼如下:
/** * 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; queue<TreeNode *> current,next; vector<int> level; if(NULL == root) return result; current.push(root); while(current.size()) { while(current.size()) { TreeNode *p; p = current.front(); current.pop(); level.push_back(p->val); if(p->left) next.push(p->left); if(p->right) next.push(p->right); } result.push_back(level); level.clear(); swap(current,next); } reverse(result.begin(),result.end()); //相對(duì)與Binary Tree Level Order Traversal只加了這一句。reverse(),反轉(zhuǎn) return result; } };
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