[Leetcode topic record-11] Determining whether a binary tree is an image (symmetric)

Symmetric tree

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

For example, this binary tree is unsupported Ric:

    1

   / \

  2   2

 / \ / \

3  4 4  3

But the following is not:

    1

   / \

  2   2

   \   \

   3    3

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

Confused what"{1,#,2,3}"Means? > Read more on how binary tree is serialized on OJ.

[Analysis 1-original] uses a recursive solution. You can check the reference for the solution.

Reference: https://oj.leetcode.com/discuss/456/recusive-solution-symmetric-optimal-solution-inordertraversal

public static boolean isSymmetric(TreeNode root) {         if(root==null) return true;         return checkIsSymmetric(root.left,root.right);    }    private static boolean checkIsSymmetric(TreeNode leftNode,TreeNode rightNode) {       if(leftNode==null&&rightNode==null) return true;       if((leftNode!=null&&rightNode==null)||(leftNode==null&&rightNode!=null)) return false;       if(leftNode.val!=rightNode.val) return false;       return checkIsSymmetric(leftNode.left,rightNode.right)&&checkIsSymmetric(leftNode.right,rightNode.left);} 

[Analysis 2-non-original] Use stack to stack left and right, and then compare each layer.

Reference: https://oj.leetcode.com/discuss/456/recusive-solution-symmetric-optimal-solution-inordertraversal

 public boolean isSymmetric(TreeNode root) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        if (root == null)            return true;        Stack<TreeNode> s1 = new Stack<TreeNode>();        Stack<TreeNode> s2 = new Stack<TreeNode>();        s1.push(root.left);        s2.push(root.right);        while (!s1.empty() && !s2.empty()) {            TreeNode n1 = s1.pop();            TreeNode n2 = s2.pop();            if (n1 == null && n2 == null)                continue;            if (n1 == null || n2 == null)                return false;            if (n1.val != n2.val)                return false;            s1.push(n1.left);            s2.push(n2.right);            s1.push(n1.right);            s2.push(n2.left);        }        return true;    }

[Leetcode topic record-11] Determining whether a binary tree is an image (symmetric)