[LeetCode]Binary Tree Level Order Traversal II
2012-11-06 15:58
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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:
... 感觉很囧在,前一题的基础上 加了一句 reverse(ret.begin(),ret.end())
.好像编程之美上有这题,没注意看。
如果是每层的也要逆序的话,就把 left 和right 入队或入栈的顺序调换一下就行了
to root).
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], ]
... 感觉很囧在,前一题的基础上 加了一句 reverse(ret.begin(),ret.end())
.好像编程之美上有这题,没注意看。
如果是每层的也要逆序的话,就把 left 和right 入队或入栈的顺序调换一下就行了
/** * Definition for binary tree * 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) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<vector<int> > ret; ret.clear(); if(root == NULL) return ret; queue<TreeNode*> S; S.push(root); S.push(NULL); vector<int> tmp; while(!S.empty()){ //travesal current level TreeNode* p = S.front(); S.pop(); if(p!=NULL) { tmp.push_back(p->val); if(p->left) S.push(p->left); if(p->right) S.push(p->right); }else{ if(!tmp.empty()) { S.push(NULL); ret.push_back(tmp); tmp.clear(); } } } reverse(ret.begin(),ret.end()); return ret; } };
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