leetcode: Symmetric Tree

判断一棵树是否对称，如果对称，中序遍历的结果一定也是对称的。

    public boolean isSymmetric(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root == null) return true;
        
        ArrayList<Integer> ret = new ArrayList<Integer>();
        inorder(root, ret);
        for(int i=0, j=ret.size()-1; i<j; i++, j--){
            if(ret.get(i) != ret.get(j))
                return false;
        }
        return true;
    }
    
    public void inorder(TreeNode root, ArrayList<Integer> ret){
        if(root == null) return;
        if(root.left!=null)
            inorder(root.left, ret);
        ret.add(root.val);
        if(root.right!=null)
            inorder(root.right, ret);
    }

看了一下别人的代码，也可以在递归的过程中直接去判断。

    public boolean isSymmetric(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root== null) return true;

        return ifSymmetric(root.left, root.right);        
    }

    public boolean ifSymmetric(TreeNode tree1, TreeNode tree2){
        if(tree1==null && tree2==null)
            return true;
        else if(tree1 == null || tree2 == null)
            return false;

        if(tree1.val != tree2.val)
            return false;
        else
            return (ifSymmetric(tree1.left, tree2.right) && ifSymmetric(tree1.right, tree2.left));
    }