101. Symmetric Tree

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.

L: mirror tree:

Node mirror(Node node)
    {
        if (node == null)
            return node;
 
        /* do the subtrees */
        Node left = mirror(node.left);
        Node right = mirror(node.right);
 
        /* swap the left and right pointers */
        node.left = right;
        node.right = left;
 
        return node;
    }

　　

Traverse both left and right branches of the root symmetricaly and check if the values are equal.

想testcase, 如何遍历, 与谁比较, 是否为空,

就是判断值和关系是否相等

先序遍历

public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        Stack<TreeNode> left = new Stack<>();
        Stack<TreeNode> right = new Stack<>();
        
        if (root.left !=  null && root.right == null || root.right != null && root.left == null) {
            return false;
        }
        if (root.left == null && root.right == null) {
            return true;
        }
        left.push(root.left);
        right.push(root.right);
        while (!left.isEmpty() && !right.isEmpty())  {
            TreeNode cur1 = left.pop();
            TreeNode cur2 = right.pop();
            if (cur1.val != cur2.val)  {
                return false;
            }
            if (cur1.left !=  null) {
                left.push(cur1.left);
            }
            if (cur2.right !=  null) {
                right.push(cur2.right);
            }
            if (left.size() != right.size()) return false;
            if (cur2.left !=  null) {
                left.push(cur2.left);
            }
            if (cur1.right !=  null) {
                right.push(cur1.right);
            }
            if (left.size() != right.size()) return false;
        
        }
        return left.size() == right.size();
    }

　　

递归

public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }
        return helper(root.left, root.right);
    }
    
    public boolean helper(TreeNode p, TreeNode q){
        if(p == null && q == null) {
            return true;
        }
        if(p == null || q == null) {
            return false;
        }
        return  (
                p.val == q.val && helper(p.left, q.right) && helper(p.right, q.left)
                );
    }