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] ]
解题思路：
此题与Binary Tree Level Order Traversal相似，只是最后的结果有一个反转。
参考 http://qiaopeng688.blog.51cto.com/3572484/1834819
代码如下：
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/**
* 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());
//相对与Binary Tree Level Order Traversal只加了这一句。reverse()，反转
return
result;
}
};
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本文转自313119992 51CTO博客，原文链接:http://blog.51cto.com/qiaopeng688/1834833
