Level Order Alternating - Amazon Top Interview Questions
Problem Statement :
Given a binary tree root, return values of the nodes in each level, alternating from going left-to-right and right-to-left. Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [3, [0, null, [2, [1, null, null], null]], [4, null, null]] Output [3, 4, 0, 2, 1]
Solution :
Solution in C++ :
vector<int> solve(Tree* root) {
vector<int> v;
stack<Tree*> st1, st2;
st1.push(root);
Tree* curr;
while (!st1.empty() or !st2.empty()) {
while (!st1.empty()) {
curr = st1.top();
st1.pop();
v.push_back(curr->val);
if (curr->left) {
st2.push(curr->left);
}
if (curr->right) {
st2.push(curr->right);
}
}
while (!st2.empty()) {
curr = st2.top();
st2.pop();
v.push_back(curr->val);
if (curr->right) {
st1.push(curr->right);
}
if (curr->left) {
st1.push(curr->left);
}
}
}
return v;
}
Solution in Python :
class Solution:
def solve(self, root):
d = deque()
d.append(root)
ans = []
lr = True
while d:
levelSize = len(d)
level = []
while levelSize:
current = d.popleft()
level.append(current.val)
if current.left:
d.append(current.left)
if current.right:
d.append(current.right)
levelSize -= 1
if lr:
ans += level
else:
ans += level[::-1]
lr = not lr
return ans
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