LeetCode 101. Symmetric Tree 对称二叉树

题目：

给定一个二叉树，检查它是否是镜像对称的。

例如，二叉树 [1,2,2,3,4,4,3] 是对称的。

    1
   / \
  2   2
 / \ / \
3  4 4  3

但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:

    1
   / \
  2   2
   \   \
   3    3

说明:

如果你可以运用递归和迭代两种方法解决这个问题，会很加分。

解题思路：

对这棵树同时进行优先访问左子树的前序遍历和优先访问右子树的前序遍历，判断当前访问到的两个节点是不是相等。

代码实现：

递归：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        TreeNode p = root;
        TreeNode q = root;
        return helper(p, q);
    }

    private boolean helper(TreeNode p, TreeNode q) {
        // 判断有没有节点为null的情况，一句话搞定
        if (p == null || q == null) return p == q;
        return (p.val == q.val) && helper(p.left, q.right) && helper(p.right, q.left);
    }
}

迭代：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        Stack<TreeNode> pStack = new Stack<TreeNode>();
        Stack<TreeNode> qStack = new Stack<TreeNode>();
        TreeNode p = root;
        TreeNode q = root;
        while (p != null && q != null) {
            // 当前节点不相等，返回false
            if (!isEqual(p, q)) return false;
            if (p.right != null) pStack.push(p.right);
            if (q.left != null) qStack.push(q.left);
            p = p.left;
            q = q.right;
            if (p == null && q == null && !pStack.isEmpty() && !qStack.isEmpty()) {
                p = pStack.pop();
                q = qStack.pop();
            }
        }
        return p == q;
    }

    private boolean isEqual(TreeNode p, TreeNode q) {
        boolean children = true;
        if (p == null || q == null) return p == q;
        if (p.left == null || q.right == null) children = children && (p.left == q.right);
        if (p.right == null || q.left == null) children = children && (p.right == q.left);
        return (p.val == q.val) && children;
    }
}