144. Binary Tree Preorder Traversal
Use iteration instead of recursion
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
//mid -> left -> right
Deque<TreeNode> stack = new LinkedList<>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode cur = stack.pop();
if(cur!=null){
list.add(cur.val);
stack.push(cur.right);
stack.push(cur.left);
}
}
return list;
}
LeetCode No. 226. Invert Binary Tree
Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:
Example 2:
Input: root = [2,1,3]
Output: [2,3,1]
Example 3:
Input: root = []
Output: []
Constraints:
The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100
BFS, Iteration ways
public TreeNode invertTree(TreeNode root) {
if(root == null){
return null;
}
TreeNode cur = root;
Deque<TreeNode> queue = new ArrayDeque<>();
queue.offer(cur);
while(!queue.isEmpty()){
if(cur!=null){
cur = queue.poll();
TreeNode temp = cur.left;
cur.left = cur.right;
cur.right = temp;
if(cur.left != null){
queue.offer(cur.left);
}
if(cur.right !=null){
queue.offer(cur.right);
}
}
}
return root;
}
Refine It
Here this evaluation is useless because if the element is in queue it must be non-null (We use offer() to add elements and it will cause NullPointer Exception it offered elements are null)
LeetCode No. 101. Symmetric Tree
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
Deque<TreeNode> leftQ = new LinkedList<>();
Deque<TreeNode> rightQ = new LinkedList<>();
leftQ.offer(root.left);
rightQ.offer(root.right);
while (!leftQ.isEmpty() && !rightQ.isEmpty()) {
TreeNode left = leftQ.poll();
TreeNode right = rightQ.poll();
if (left == null && right == null) {
continue;
}
if (left == null || right == null) {
return false;
}
if (left.val != right.val) {
return false;
}
leftQ.offerLast(left.left);
leftQ.offerLast(left.right);
rightQ.offerLast(right.right);
rightQ.offerLast(right.left);
}
return leftQ.isEmpty() && rightQ.isEmpty();
}
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