package com.hit.basmath.learn.binary_tree;

import com.hit.common.TreeNode;

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

/**
 * 101. Symmetric Tree
 * <p>
 * Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
 * <p>
 * For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
 * <p>
 * 1
 * / \
 * 2   2
 * / \ / \
 * 3  4 4  3
 * <p>
 * But the following [1,2,2,null,3,null,3] is not:
 * <p>
 * 1
 * / \
 * 2   2
 * \   \
 * 3    3
 * <p>
 * Note:
 * <p>
 * Bonus points if you could solve it both recursively and iteratively.
 */
public class _101 {

    /**
     * recursive method
     *
     * @param root pending node
     * @return true or false
     */
    public boolean isSymmetric(TreeNode root) {
        return root == null || isSymmetricHelper(root.left, root.right);
    }

    private boolean isSymmetricHelper(TreeNode left, TreeNode right) {
        if (left == null || right == null) return left == right;
        return left.val == right.val
                && isSymmetricHelper(left.left, right.right)
                && isSymmetricHelper(left.right, right.left);

    }


    /**
     * Non-recursive method
     *
     * @param root pending node
     * @return true or false
     */
    public boolean isSymmetric2(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        queue.add(root);
        while (!queue.isEmpty()) {
            TreeNode t1 = queue.poll();
            TreeNode t2 = queue.poll();
            if (t1 == null && t2 == null) continue;
            if (t1 == null || t2 == null) return false;
            if (t1.val != t2.val) return false;
            queue.add(t1.left);
            queue.add(t2.right);
            queue.add(t1.right);
            queue.add(t2.left);
        }
        return true;
    }
}