題目：請實現一個函式來判斷一棵二叉樹是不是對稱的。如果一棵二叉樹和它的映象一樣，那麼它是對稱的。

解題思路

　　通常我們有三種不同的二叉樹遍歷演算法，即前序遍歷、中序遍歷和後序遍歷。在這三種遍歷演算法中，都是先遍歷左子結點再遍歷右子結點。我們是否可以定義一種遍歷演算法，先遍歷右子結點再遍歷左子結點？比如我們針對前序遍歷定義一種對稱的遍歷演算法，即先遍歷父節點，再遍歷它的右子結點，最後遍歷它的左子結點。
　　我們發現可以用過比較二叉樹的前序遍歷序列和對稱前序遍歷序列來判斷二叉樹是不是對稱的。如果兩個序列一樣，那麼二叉樹就是對稱的。

結點定義

private static class
 
 BinaryTreeNode {
    private int val;
    private BinaryTreeNode left;
    private BinaryTreeNode right;

    public BinaryTreeNode() {
    }

    public BinaryTreeNode(int val) {
        this.val = val;
    }

    @Override
    public String toString() {
        return val + "";
    }
}

程式碼實現

public
 
 class Test59 {
    private static class BinaryTreeNode {
        private int val;
        private BinaryTreeNode left;
        private BinaryTreeNode right;

        public BinaryTreeNode() {
        }

        public BinaryTreeNode(int val) {
            this.val = val;
        }

        @Override
        public
 
 String toString() {
            return val + "";
        }
    }

    public static boolean isSymmetrical(BinaryTreeNode root) {
        return isSymmetrical(root, root);
    }

    private static boolean isSymmetrical(BinaryTreeNode left, BinaryTreeNode right) {

        if (left == null && right == null) {
            return true;
        }

        if (left == null || right == null) {
            return false;
        }

        if (left.val != right.val ) {
            return false;
        }

        return isSymmetrical(left.left, right.right) && isSymmetrical(left.right, right.left);
    }

    public static void main(String[] args) {
        test01();
        test02();
    }

    private static void assemble(BinaryTreeNode node,
                                 BinaryTreeNode left,
                                 BinaryTreeNode right) {
        node.left = left;
        node.right = right;
    }

    //                            1
    //                  2                   2
    //             4         6          6          4
    //          8     9   10   11   11     10   9     8
    public static void test01() {
        BinaryTreeNode n1 = new BinaryTreeNode(1);
        BinaryTreeNode n2 = new BinaryTreeNode(2);
        BinaryTreeNode n3 = new BinaryTreeNode(2);
        BinaryTreeNode n4 = new BinaryTreeNode(4);
        BinaryTreeNode n5 = new BinaryTreeNode(6);
        BinaryTreeNode n6 = new BinaryTreeNode(6);
        BinaryTreeNode n7 = new BinaryTreeNode(4);
        BinaryTreeNode n8 = new BinaryTreeNode(8);
        BinaryTreeNode n9 = new BinaryTreeNode(9);
        BinaryTreeNode n10 = new BinaryTreeNode(10);
        BinaryTreeNode n11 = new BinaryTreeNode(11);
        BinaryTreeNode n12 = new BinaryTreeNode(11);
        BinaryTreeNode n13 = new BinaryTreeNode(10);
        BinaryTreeNode n14 = new BinaryTreeNode(9);
        BinaryTreeNode n15 = new BinaryTreeNode(8);

        assemble(n1, n2, n3);
        assemble(n2, n4, n5);
        assemble(n3, n6, n7);
        assemble(n4, n8, n9);
        assemble(n5, n10, n11);
        assemble(n6, n12, n13);
        assemble(n7, n14, n15);
        assemble(n8, null, null);
        assemble(n9, null, null);
        assemble(n10, null, null);
        assemble(n11, null, null);
        assemble(n12, null, null);
        assemble(n13, null, null);
        assemble(n14, null, null);
        assemble(n15, null, null);

        System.out.println(isSymmetrical(n1));

    }


    //                            1
    //                  2                   2
    //             4         5          6          4
    //          8     9   10   11   11     10   9     8
    public static void test02() {
        BinaryTreeNode n1 = new BinaryTreeNode(1);
        BinaryTreeNode n2 = new BinaryTreeNode(2);
        BinaryTreeNode n3 = new BinaryTreeNode(2);
        BinaryTreeNode n4 = new BinaryTreeNode(4);
        BinaryTreeNode n5 = new BinaryTreeNode(5);
        BinaryTreeNode n6 = new BinaryTreeNode(6);
        BinaryTreeNode n7 = new BinaryTreeNode(4);
        BinaryTreeNode n8 = new BinaryTreeNode(8);
        BinaryTreeNode n9 = new BinaryTreeNode(9);
        BinaryTreeNode n10 = new BinaryTreeNode(10);
        BinaryTreeNode n11 = new BinaryTreeNode(11);
        BinaryTreeNode n12 = new BinaryTreeNode(11);
        BinaryTreeNode n13 = new BinaryTreeNode(10);
        BinaryTreeNode n14 = new BinaryTreeNode(9);
        BinaryTreeNode n15 = new BinaryTreeNode(8);

        assemble(n1, n2, n3);
        assemble(n2, n4, n5);
        assemble(n3, n6, n7);
        assemble(n4, n8, n9);
        assemble(n5, n10, n11);
        assemble(n6, n12, n13);
        assemble(n7, n14, n15);
        assemble(n8, null, null);
        assemble(n9, null, null);
        assemble(n10, null, null);
        assemble(n11, null, null);
        assemble(n12, null, null);
        assemble(n13, null, null);
        assemble(n14, null, null);
        assemble(n15, null, null);

        System.out.println(isSymmetrical(n1));

    }
}

執行結果