[Leetcode 95] 103 Binary Tree Zigzag Level Order Traversal

Problem:

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]
]

Analysis:

A little change to the original level-order reaversal code. Add a zz flag to indicate whether or not reverse the current level's elements.

Remember that after exit the while loop, an extra push_back operation is needed !

Code:

/**
 * 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> > res;
        
        if (root == NULL)
            return res;
        
        vector<int> l;
        
        queue<TreeNode*> q;
        q.push(root);
        
        TreeNode *senti = new TreeNode(-1);
        senti->left = senti->right = NULL;
        q.push(senti);
        
        bool zz = false;
        
        while (q.size() != 1) {
            TreeNode *n = q.front();
            q.pop();
            
            if (n == senti) { // get the sentinal node
                if (zz) reverse(l.begin(), l.end());
                
                res.push_back(l);
                q.push(n);
                zz = !zz;
                l.clear();
            } else {
                l.push_back(n->val);
            
                if (n->left != NULL) q.push(n->left);
                if (n->right != NULL) q.push(n->right);
            }
        }
        if (zz) reverse(l.begin(), l.end());
        res.push_back(l);
        
        return res;
    }
};