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] ]
同上一篇,只是多做了一次reverse: http://blog.youkuaiyun.com/anyicheng2015/article/details/42264721
/**
* 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) {
vector<vector<int> > results;
if(root == NULL) return results;
queue<TreeNode *> onelevel;
onelevel.push(root);
bool shouldReverse = false;
while(!onelevel.empty()) {
// Get a level and get TreeNode of the next level.
vector<int> valOfCurrentLevel;
queue<TreeNode *> nextlevel;
while(!onelevel.empty()) {
TreeNode *front = onelevel.front();
valOfCurrentLevel.push_back(front->val);
if(front->left != NULL)
nextlevel.push(front->left);
if(front->right != NULL)
nextlevel.push(front->right);
onelevel.pop();
}
if(shouldReverse)
reverse(valOfCurrentLevel.begin(), valOfCurrentLevel.end());
results.push_back(valOfCurrentLevel);
onelevel = nextlevel;
shouldReverse = !shouldReverse;
}
return results;
}
};