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Leetcode bfs&dfs Binary Tree Postorder Traversal II
Binary Tree Level Order Traversal II
Total Accepted: 16983 Total Submissions: 54229My SubmissionsGiven 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
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
题意：从底往上按层遍历二叉树
思路：
思路和Binary Tree Level Order Traveral 一样，
即从上往下按层遍历二叉树，将每一层的节点存放到该层对应的数组中
最后将得到的数组倒转一下就可以了
按层遍历二叉树可用bfs，也可用dfs，但都要记录节点所在的层
复杂度：时间O(n)， 空间O(n)
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void levelOrderBottom(TreeNode *root, int level, vector<vector<int> >&result)
{
if(!root) return ;
if(level >= result.size()) {
vector<int> temp;
temp.push_back(root->val);
result.push_back(temp);
}
else result[level].push_back(root->val);
levelOrderBottom(root->left, level + 1, result);
levelOrderBottom(root->right, level + 1, result);
}
vector<vector<int> > levelOrderBottom(TreeNode *root){
vector<vector<int> >result;
levelOrderBottom(root, 0, result);
reverse(result.begin(), result.end());
return result;
}
};Leetcode bfs&dfs Binary Tree Postorder Traversal II
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