public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
Deque<TreeNode> queue = new LinkedList<>();
if (root == null){
return result;
}
queue.add(root);
while(!queue.isEmpty()){
List<Integer> level = new ArrayList<>();
// queue size is changing is for loop
int level_size = queue.size();
for (int i = 0; i < level_size; i++){
TreeNode node = queue.poll();
level.add(node.val);
if (node.left != null){
queue.add(node.left);
}
if (node.right != null){
queue.add(node.right);
}
}
result.add(level);
}
return result;
}
107 Level traverse - bottom up
top down then reverse
get the depth first, use dfs to store elements in each level
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null){
return result;
}
int maxLevel = depth(root);
int curLevel = 1;
while(curLevel <= maxLevel){
List<Integer> level = new ArrayList<>();
dfs(root, curLevel, maxLevel, level);
result.add(level);
maxLevel--;
}
return result;
}
private int depth(TreeNode root){
if (root == null){
return 0;
}
int left = depth(root.left);
int right = depth(root.right);
return 1 + (left > right ? left : right);
}
private void dfs(TreeNode root, int curLevel, int maxLevel, List<Integer> level){
if (root == null){
return ;
}
if (curLevel == maxLevel){
level.add(root.val);
}
dfs(root.left, curLevel + 1, maxLevel, level);
dfs(root.right, curLevel + 1, maxLevel, level);
}
103 ZigZag
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).
Add array to result
Use % to decide odd or even array
ArrayList.add(0, int) - addFirst
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null){
return result;
}
preorder(root, 0, result);
return result;
}
private void preorder(TreeNode root, int level, List<List<Integer>> result){
if (root == null){
return;
}
if (result.size() < level + 1){
result.add(new ArrayList<>());
}
List<Integer> tmp = result.get(level);
if (level % 2 == 0){
tmp.add(root.val);
}else{
tmp.add(0, root.val); // addFirst
}
preorder(root.left, level + 1, result);
preorder(root.right, level + 1, result);
}
297. Serialize and Deserialize Binary Tree
output string is preorder string
def serialize(self, root):
s = ''
if root is None: s += '#' + ','
else:
s += str(root.val) + ','
s += self.serialize(root.left)
s += self.serialize(root.right)
return s
def deserialize(self, data):
q = data.split(',')
return self.dHelper(q)
def dHelper(self, q):
val = q.pop(0)
if val == '#': return None
else:
node = TreeNode(val)
node.left = self.dHelper(q)
node.right = self.dHelper(q)
return node