LeetCode

题目：

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

For example, this binary tree is symmetric:

    1
   / 
  2   2
 /  / 
3  4 4  3

But the following is not:

    1
   / 
  2   2
      
   3    3

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

思路：

1) 递归

package tree;

public class SymmetricTree {

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

}

2) 迭代

package tree;

import java.util.Stack;

public class SymmetricTree {

    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        Stack<TreeNode> stack1 = new Stack<TreeNode>();
        Stack<TreeNode> stack2 = new Stack<TreeNode>();
        TreeNode p = root.left;
        TreeNode q = root.right;
        while ((p != null && q != null) || (!stack1.isEmpty() && !stack2.isEmpty())) {
            while (p != null && q != null) {
                stack1.push(p);
                p = p.left;
                stack2.push(q);
                q = q.right;
            } 
            
            if (p != null || q != null) return false;
            
            if (stack1.isEmpty() && stack1.isEmpty()) return true;
            if (stack1.isEmpty() || stack1.isEmpty()) return false;
            
            TreeNode node1 = stack1.pop();
            TreeNode node2 = stack2.pop();
            if (node1.val != node2.val) return false;
            p = node1.right;
            q = node2.left;            
        }
        
        return p == null && q == null && stack1.isEmpty() && stack2.isEmpty();
    }
    
}