leetcode:Binary Tree Zigzag Level Order Traversal

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},

return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
分析
Assuming after traversing the 1st level, nodes in queue are {9, 20, 8},
And we are going to traverse 2nd level, which is even line and should print value from right to left [8, 20, 9].

We know there are 3 nodes in current queue,
so the vector for this level in final result should be of size 3.
Then, queue [i] -> goes to -> vector[queue.size() - 1 - i] i.e.
the ith node in current queue should be placed in (queue.size() - 1 - i) position in vector for that line.

For example, for node(9), it’s index in queue is 0, so its index in vector should be (3-1-0) = 2.
代码

vector<vector<int> > zigzagLevelOrder(TreeNode* root) {
    if (root == NULL) {
        return vector<vector<int> > ();
    }
    vector<vector<int> > result;

    queue<TreeNode*> nodesQueue;
    nodesQueue.push(root);
    bool leftToRight = true;

    while ( !nodesQueue.empty()) {
        int size = nodesQueue.size();
        vector<int> row(size);
        for (int i = 0; i < size; i++) {
            TreeNode* node = nodesQueue.front();
            nodesQueue.pop();

            // find position to fill node‘s value
            int index = (leftToRight) ? i : (size - 1 - i);

            row[index] = node->val;
            if (node->left) {
                nodesQueue.push(node->left);
            }
            if (node->right) {
                nodesQueue.push(node->right);
            }
        }
        // after this level
        leftToRight = !leftToRight;
        result.push_back(row);
    }
    return result;
}