Binary Tree Zigzag Level Order Traversal

Q:

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]
]

A:

注意方向即可

/**
 * 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> > results;
        vector<int> result;
        if (!root) return results;
        vector<TreeNode*> cur_nodes;
        cur_nodes.push_back(root);
        result.push_back(root->val);
        results.push_back(result);
        bool reverse = true;
        while (!cur_nodes.empty()) {
            result.clear();
            vector<TreeNode*> next_nodes;
            for (int i = cur_nodes.size() - 1; i >= 0; --i) {
                if (reverse) {
                    if (cur_nodes[i]->right) {
                        next_nodes.push_back(cur_nodes[i]->right);
                        result.push_back(cur_nodes[i]->right->val);
                    }
                    if (cur_nodes[i]->left) {
                        next_nodes.push_back(cur_nodes[i]->left);
                        result.push_back(cur_nodes[i]->left->val);
                    }
                } else {
                    if (cur_nodes[i]->left) {
                        next_nodes.push_back(cur_nodes[i]->left);
                        result.push_back(cur_nodes[i]->left->val);
                    }
                    if (cur_nodes[i]->right) {
                        next_nodes.push_back(cur_nodes[i]->right);
                        result.push_back(cur_nodes[i]->right->val);
                    }
                }
            }
            reverse = !reverse;
            if (result.empty()) break;
            results.push_back(result);
            cur_nodes.swap(next_nodes);
        }
        return results;
    }
};