L101. Symmetric Tree

binary tree

Problem:

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

Solution:

• if it is null, it is mirrored, otherwise , compare left and right's value.

public boolean isSymmetric(TreeNode root){
    return root == null ? true : isMirror(root.left, root.right);
}
private boolean isMirror(TreeNode left, TreeNode right){
    if(left == null && right == null) return true;
    if(left == null || right == null) return false;
    if(left.val != right.val) return false;
    return isMirror(left.left, right.right) && isMirror(left.right, right.left);
}

• Iteration with two stack

public boolean isSymmetric(TreeNode root) {
  if (root == null) {
    return true;
  }
  if (root.left == null && root.right == null) return true;
  if (root.left == null || root.right == null) return false;
  // children are not null
  Stack<TreeNode> stack = new Stack<>();
  stack.push(root.left);
  stack.push(root.right);
  
  while (stack.size() > 0) {
    TreeNode t1 = stack.pop();
    TreeNode t2 = stack.pop();
    // null check
    if (t1 == null && t2 == null) continue;
    if (t1 == null || t2 == null) return false;
    // value check
    if (t1.val != t2.val) return false;
    // push children
    stack.push(t1.right); stack.push(t2.left); // could be null
    stack.push(t1.left); stack.push(t2.right);
  }
  
  return true;
}