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levelOrderBottom.cpp
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levelOrderBottom.cpp
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// https://leetcode.com/problems/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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
// 把上一题(层序遍历)的输出结果翻转一下然后输入。。
// 不过我怀疑出题者不是这个意思。。
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
queue<TreeNode *> parent;
queue<TreeNode *> child;
vector<vector<int>> vvi;
vector<int> buff; // 临时存放当前层的值
auto *ptr_parent = &parent;
auto *ptr_child = &child;
if(root == nullptr)
return vvi;
ptr_parent->push(root);
while(true){
buff.clear();
while(!ptr_parent->empty()){
auto front = ptr_parent->front();
ptr_parent->pop();
if(front->left)
ptr_child->push(front->left);
if(front->right)
ptr_child->push(front->right);
buff.push_back(front>val);
}
if(!buff.empty())
vvi.push_back(buff);
if(ptr_child->empty())
break;
// 交换两个指针
auto tmp=ptr_parent;
ptr_parent = ptr_child;
ptr_child = tmp;
}
reverse(res.begin(), res.end()); // 要包含头文件<algorithm>
return vvi;
}
};