101. Symmetric Tree

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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

Note: * Bonus points if you could solve it both recursively and iteratively.

分析

二叉树对称，iff, 左子树和右子树对称。 那么怎么判断两棵树对称？

两棵树对成，iff

• 左节点和右节点的值相等
• 两棵树的左子树和右子树依次对称

public boolean isSymmetricSubTree(TreeNode root) {
    if (root == null) return true;
    return isSymmetricNodes(root.left, root.right);
}

private boolean isSymmetricSubTree(TreeNode root1, TreeNode root2) {
    if (root1 == null || root2 ==  null) return root1 == root2;
    return root1.val == root2.val && 
                isSymmetricSubTree(root1.left, root2.right) &&
                isSymmetricSubTree(root1.right, root2.left);
}

isSymmetricSubTree还可以再优化一点, 也就是将根节点看成是子节点，相当于增加了一个虚拟根节点。

public boolean isSymmetricSubTree(TreeNode root) {
    return isSymmetricNodes(root, root);
}

同样的思路，使用迭代:

public boolean isSymmetric(TreeNode root) {
    if (root == null) return true;
    Stack<TreeNode> stack = new Stack<>();
    stack.push(root);
    stack.push(root);
    while (!stack.isEmpty()) {
        TreeNode left = stack.pop();
        TreeNode right = stack.pop();
        if (left.val != right.val) return false;
        if (left.left != null && right.right != null) {
            stack.push(left.left); stack.push(right.right);
        } else if (left.left != right.right) return false;

        if (left.right != null && right.left != null) {
            stack.push(left.right); stack.push(right.left);
        } else if  (left.right != right.left) return false;
    }
    return true;
}

发现使用BFS思路的QUEUE，使用迭代可以简化一些：

public boolean isSymmetric(TreeNode root) {
    Queue<TreeNode> q = new LinkedList<>();
    q.add(root);
    q.add(root);
    while (!q.isEmpty()) {
        TreeNode t1 = q.poll();
        TreeNode t2 = q.poll();
        if (t1 == null && t2 == null) continue;
        if (t1 == null || t2 == null) return false;
        if (t1.val != t2.val) return false;
        q.add(t1.left);
        q.add(t2.right);
        q.add(t1.right);
        q.add(t2.left);
    }
    return true;
}