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;
}