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Binary-Tree-Level-Order-Traversal-II.cpp
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Binary-Tree-Level-Order-Traversal-II.cpp
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/*
* Song Qiang <qiang.song@usc.edu> 2013
*
*/
/*
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], ]
confused what "{1,#,2,3}" means? > read more on how binary tree is
serialized on OJ.
*/
#include <cmath>
#include <cassert>
#include <vector>
#include <numeric>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <limits>
#include <utility>
#include <set>
#include <queue>
using namespace std;
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (root == NULL) return vector<vector<int> >();
queue<TreeNode*> q;
q.push(root);
q.push(NULL);
vector<vector<int> > results;
vector<int> r;
while (!q.empty())
{
const TreeNode* p = q.front();
q.pop();
if (p == NULL)
{
if (!r.empty()) results.push_back(r);
if (!q.empty()) q.push(NULL);
r.clear();
}
else
{
r.push_back(p->val);
if (p->left != NULL) q.push(p->left);
if (p->right != NULL) q.push(p->right);
}
}
reverse(results.begin(), results.end());
return results;
}
};
int
main(int argn, char** argv)
{
Solution s;
}