算法设计与应用基础系列10



103. 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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

[  [3],  [20,9],  [15,7]]

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.

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;}