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107-BinaryTreeLevelOrderTraversalII.cpp
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107-BinaryTreeLevelOrderTraversalII.cpp
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/*
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) {}
* };
*/
#include <stdio.h>
#include <vector>
using namespace std;
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;
if (root == NULL) {
return result;
}
vector<TreeNode *> store;
store.push_back(root);
int visited = 0;
int lastlevel = 0;
while (visited < store.size()) {
lastlevel = store.size();
vector<int> level;
while (visited < lastlevel) {
level.push_back(store[visited]->val);
printf("Visit %d\n", store[visited]->val);
if (store[visited]->left != NULL) {
store.push_back(store[visited]->left);
}
if (store[visited]->right != NULL) {
store.push_back(store[visited]->right);
}
visited++;
}
result.push_back(level);
}
vector <vector <int> > ret;
for (size_t i = 0; i < result.size(); i++) {
ret.push_back(result[result.size() - 1 - i]);
}
return ret;
}
};
int main() {
TreeNode t1(1), t2(2), t3(3), t4(4), t5(5), t6(6), t7(7), t8(8), t9(9), t10(10);
t4.left = &t2;
t4.right = &t5;
t2.left = &t1;
t2.right = &t3;
t5.right = &t9;
t5.left = &t10;
t9.left = &t7;
t7.left = &t6;
t7.right = &t8;
Solution s;
vector <vector<int> > ret = s.levelOrderBottom(&t4);
for (size_t i = 0; i < ret.size(); i++) {
for (size_t j = 0; j < ret[i].size(); j++) {
printf("%d ", ret[i][j]);
}
printf("\n");
}
return 0;
}