Leetcode之Binary Tree Zigzag Level Order Traversal

题目：

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]
]

代码：

/**
 * 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)return res;
	stack<TreeNode*> s;
	s.push(root);
	while (!s.empty()) {
		vector<int> v1; vector<TreeNode*> v;
		while (!s.empty()) {
			v1.push_back(s.top()->val);
			v.push_back(s.top());
			s.pop();
		}
		if(v1.size()>0)res.push_back(v1);
		for (int i = 0; i < v.size(); i++) {
			if (v[i]->left)s.push(v[i]->left);
			if (v[i]->right)s.push(v[i]->right);
		}

		vector<int> v2; vector<TreeNode*> vt;
		while (!s.empty()) {
			v2.push_back(s.top()->val);
			vt.push_back(s.top());
			s.pop();
		}
		if(v2.size()>0)res.push_back(v2);
		for (int i = 0; i < vt.size(); i++) {
			if (vt[i]->right)s.push(vt[i]->right);
			if (vt[i]->left)s.push(vt[i]->left);
		}
	}

	return res;
    }
};

想法：

stack