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[leetcode 107]Binary Tree Level Order Traversal II

2015-11-01 18:02 351 查看
/**
* 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 {
private:
int heightOfTree(TreeNode* root){
if(root == NULL)
return 0;
else
return max(heightOfTree(root->left),heightOfTree(root->right))+1;

}
public:
//用队列的方法
/* vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> result;
vector<int> vals;
queue<TreeNode*> que;
if(root == NULL)
return result;
que.push(root);
que.push(NULL);//用每一层结束入队后加入NULL来识别层数;
while(!que.empty()){
TreeNode* node;
node = que.front();
que.pop();
if(node != NULL){
vals.push_back(node->val);
if(node->left != NULL)
que.push(node->left);
if(node->right != NULL)
que.push(node->right);
}
else{
if(!vals.empty()){
result.push_back(vals);
vals.clear();
que.push(NULL);
}
}
}
reverse(result.begin(),result.end());
return result;
}*/
//记录每层层数的方法
void levelOrder(vector<vector<int>> &ans, TreeNode *node, int level) {
if (!node) return;
ans[level].push_back(node->val);
levelOrder(ans,node->left,level-1);
levelOrder(ans,node->right,level-1);
}

vector<vector<int>> levelOrderBottom(TreeNode* root) {
int d = heightOfTree(root);
vector<vector<int>> ans(d);
levelOrder(ans,root,d-1);
return ans;
}
};
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