leetcode151: symmetric tree

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

給定一顆二叉樹，判斷該二叉樹是否是對稱的。

For example, this binary tree is symmetric:

如下面的二叉樹是對稱的。

        1

       / \

      2   2

     /     \

    3       3

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

提示：是否可以同時用遞迴和迭代的方法解決？

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方法一：遞迴

public class SymmetricTree {
    static class TreeNode {
           int val;
            TreeNode left;
            TreeNode right;
            TreeNode(int x) { val = x; }
        }
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return isSymmetric(root.left, root.right);
    }
    public boolean isSymmetric(TreeNode leftNode, TreeNode rightNode){
        if(leftNode == null && rightNode == null ) return true;
        if(leftNode == null || rightNode == null) return false;
        if(leftNode.val != rightNode.val) return  false;
        boolean isLeft = isSymmetric(leftNode.left, rightNode.right);
        boolean isRight = isSymmetric(leftNode.right, rightNode.left);
        return isLeft && isRight;
    }

    public static void main(String args[]){
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(2);
        root.left.left = new TreeNode(3);
        root.left.right = new TreeNode(4);
        root.right.left = new TreeNode(4);
        root.right.right = new TreeNode(3);
        System.out.println(new SymmetricTree().isSymmetric(root));
        System.out.println(new SymmetricTree().isSymmetricIter(root));
    }
}

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方法二：迭代

public boolean isSymmetricIter(TreeNode root) {
        if(root == null) return true;
        Stack<TreeNode> leftStack = new Stack<TreeNode>();
        Stack<TreeNode> rightStack = new Stack<TreeNode>();
        leftStack.push(root.left);
        rightStack.push(root.right);

        while (leftStack.size() > 0 && rightStack.size() > 0){
            TreeNode left = leftStack.pop();
            TreeNode right = rightStack.pop();
            if(left == null && right == null) continue;
            if(left == null || right == null) return false;
            if(left.val == right.val) {
                leftStack.push(left.right);
                leftStack.push(left.left);
                rightStack.push(right.left);
                rightStack.push(right.right);
            }else{
                return false;
            }
        }
        return true;
    }