# Symmetric Tree

### 描述​

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.

### 递归版​

// Symmetric Tree// 递归版，时间复杂度O(n)，空间复杂度O(logn)public class Solution {    public boolean isSymmetric(TreeNode root) {        if (root == null) return true;        return isSymmetric(root.left, root.right);    }    private static boolean isSymmetric(TreeNode p, TreeNode q) {        if (p == null && q == null) return true;   // 终止条件        if (p == null || q == null) return false;  // 终止条件        return p.val == q.val      // 三方合并                && isSymmetric(p.left, q.right)                && isSymmetric(p.right, q.left);    }}

### 迭代版​

// Symmetric Tree// 迭代版，时间复杂度O(n)，空间复杂度O(logn)public class Solution {    public boolean isSymmetric (TreeNode root) {        if (root == null) return true;        Stack<TreeNode> s = new Stack<>();        s.push(root.left);        s.push(root.right);        while (!s.isEmpty()) {            TreeNode p = s.pop ();            TreeNode q = s.pop ();            if (p == null && q == null) continue;            if (p == null || q == null) return false;            if (p.val != q.val) return false;            s.push(p.left);            s.push(q.right);            s.push(p.right);            s.push(q.left);        }        return true;    }}