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lc-103.cpp
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lc-103.cpp
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/**
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
* 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>> zigzagLevelOrder(TreeNode* root) {
stack<TreeNode*> currentLevel;
stack<TreeNode*> nextLevel;
TreeNode* current;
currentLevel.push(root);
int levelIndex = 0;
int l2r = 1;
vector<vector<int>> output;
if (root) output.push_back(vector<int>());
while (!currentLevel.empty()){
current = currentLevel.top();
currentLevel.pop();
if (current == NULL){
continue;
}
cout<<levelIndex<<" "<<current->val<<endl;
output[levelIndex].push_back(current->val);
if (l2r>0){
if (current->left != NULL) nextLevel.push(current->left);
if (current->right != NULL) nextLevel.push(current->right);
} else {
if (current->right != NULL) nextLevel.push(current->right);
if (current->left != NULL) nextLevel.push(current->left);
}
if (currentLevel.empty() && !nextLevel.empty()){
levelIndex++;
output.push_back(vector<int>());
swap(currentLevel, nextLevel);
l2r *= -1;
}
}
return output;
}
};