101. Symmetric Tree

Q

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:

But the following [1,2,2,null,3,null,3] is not:

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

A

recursively(C)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 bool judge(struct TreeNode *left, struct TreeNode *right) {
     if (left == NULL && right == NULL) {
         return true;
     }
     if (left == NULL || right == NULL) {
         return false;
     }
     if (left->val != right->val) {
         return false;
     }

     return (judge(left->left, right->right) && judge(left->right, right->left));
 }

bool isSymmetric(struct TreeNode* root) {
    if (root == NULL) {
        return true;
    }

    return judge(root->left, root->right);
}

iteratively(C++)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if (root == NULL) {
            return true;
        }
        if (root->left == NULL && root->right == NULL) {
            return true;
        }
        if (root->left == NULL || root->right == NULL) {
            return false;
        }

        queue<TreeNode *> q1;
        queue<TreeNode *> q2;
        q1.push(root->left);
        q2.push(root->right);
        // 每个队列层序遍历
        while (!q1.empty() && !q2.empty()) {
            TreeNode *l1, *l2;
            l1 = q1.front();
            l2 = q2.front();
            if (l1->val != l2->val) {
                return false;
            }

            if ((l1->left==NULL&&l2->right!=NULL) || (l1->left!=NULL&&l2->right==NULL)) {
                return false;
            }
            if ((l1->right==NULL&&l2->left!=NULL) || (l1->right!=NULL&&l2->left==NULL)) {
                return false;
            }

            if (l1->left!=NULL && l2->right!=NULL) {
                q1.push(l1->left);
                q2.push(l2->right);
            }
            if (l1->right!=NULL && l2->left!=NULL) {
                q1.push(l1->right);
                q2.push(l2->left);
            }
            q1.pop();
            q2.pop();
        }

        return true;
    }
};