LeetCode——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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

» Solve this problem

Binary Tree Level Order Traversal改改就o了。

/**
 * 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> > answer;
        if (root != NULL) {
            vector<TreeNode *> pre, cur;
            int dir = 1;
            pre.push_back(root);            
            while (!pre.empty()) {
                vector<int> temp;
                if (dir > 0) {
                    for (int i = 0; i < pre.size(); i++) {
                        temp.push_back(pre[i]->val);
                    }
                }
                else {
                    for (int i = pre.size() - 1; i >= 0; i--) {
                        temp.push_back(pre[i]->val);
                    }
                }
                answer.push_back(temp);
                dir = -dir;
                
                cur.clear();
                for (vector<TreeNode *>::iterator iter = pre.begin(); iter != pre.end(); iter++) {
                    if ((*iter)->left != NULL) {
                        cur.push_back((*iter)->left);
                    }
                    if ((*iter)->right != NULL) {
                        cur.push_back((*iter)->right);                        
                    }
                }                
                pre = cur;
            }
        }        
        return answer;
    }
};