LeetCode Q103: 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 {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

Solution: Similar to Q102, but need a flag to indicate whether to reverse accessed nodes at current layer.

	/**
	* 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) {
	vector<vector<int> > res;
	if(root==NULL)
	return res;
	queue<TreeNode*> Q;
	Q.push(root);
	int marker=0;
	int cur=1;
	int next=0;
	vector<int> tmp;
	while(!Q.empty()){
	TreeNode* front = Q.front();
	Q.pop();
	if(front!=NULL){
	tmp.push_back(front->val);
	Q.push(front->left);
	Q.push(front->right);
	next+=2;
	}
	cur--;
	if(cur==0&&!tmp.empty()){
	if(marker==1)
	reverse(tmp.begin(), tmp.end());
	res.push_back(tmp);
	tmp.clear();
	cur=next;
	next=0;
	marker=marker^1;
	}
	}
	return res;
	}
	};

view raw Q103.cpp hosted with ❤ by GitHub