/*
Binary Tree Zigzag Level Order Traversal
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
return its zigzag level order traversal as:
*/For example:
Given binary tree
{3,9,20,#,#,15,7}, 3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
/**
* Definition for binary tree
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int>> result;
vector<int> line;
if (root==NULL) return result;
queue<TreeNode*> q;
int leftValue = 2000;
TreeNode *leftFirst= new TreeNode(leftValue);
int rightValue = -2000;
TreeNode *rightFirst = new TreeNode(rightValue);
TreeNode *curr;
q.push(root);
q.push(leftFirst);
bool fromLeft = true;
while(!q.empty()){
curr=q.front();
q.pop();
if(curr!=leftFirst && curr!=rightFirst){
if(curr->left) q.push(curr->left);
if(curr->right) q.push(curr->right);
if(fromLeft) line.push_back(curr->val);
else line.insert(line.begin(),curr->val);
}
if(curr==rightFirst){
fromLeft = !fromLeft;
result.push_back(line);
line.clear();
if((!q.empty()) && q.front()!=rightFirst && q.front()!=leftFirst){
q.push(leftFirst);
}
}
if(curr==leftFirst){
fromLeft = !fromLeft;
result.push_back(line);
line.clear();
if((!q.empty()) && q.front()!=rightFirst && q.front()!=leftFirst){
q.push(rightFirst);
}
}
}
return result;
}
};
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