LeetCode: Symmetric Tree

看了网上答案

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool dfs(TreeNode *left, TreeNode *right) {
        if (!left && !right) return true;
        if (!left || !right) return false;
        return left->val == right->val && dfs(left->left, right->right) && dfs(left->right, right->left);
    }
    bool isSymmetric(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!root) return true;
        return dfs(root->left, root->right);
    }
};

 下面是自己写的iterasive的法子

/**
 * Definition for binary tree
 * 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) return true;
        queue<TreeNode*> ltr, rtl;
        ltr.push(root), rtl.push(root);
        while (!ltr.empty() && !rtl.empty()) {
            TreeNode *ltrfront = ltr.front();
            TreeNode *rtlfront = rtl.front();
            ltr.pop(), rtl.pop();
            if (!ltrfront && !rtlfront) continue;
            if (!ltrfront || !rtlfront || ltrfront->val != rtlfront->val) return false;
            ltr.push(ltrfront->left), ltr.push(ltrfront->right);
            rtl.push(rtlfront->right), rtl.push(rtlfront->left);
        }
        return ltr.empty() && rtl.empty();
    }
};

 C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public bool IsSymmetric(TreeNode root) {
        if (root == null) return true;
        Queue<TreeNode> ltr = new Queue<TreeNode>();
        Queue<TreeNode> rtl = new Queue<TreeNode>();
        ltr.Enqueue(root);
        rtl.Enqueue(root);
        while (ltr.Count > 0 && rtl.Count > 0) {
            TreeNode ltrPeek = ltr.Peek();
            TreeNode rtlPeek = rtl.Peek();
            ltr.Dequeue();
            rtl.Dequeue();
            if (ltrPeek == null && rtlPeek == null) continue;
            if (ltrPeek == null || rtlPeek == null || ltrPeek.val != rtlPeek.val) return false;
            ltr.Enqueue(ltrPeek.left);
            ltr.Enqueue(ltrPeek.right);
            rtl.Enqueue(rtlPeek.right);
            rtl.Enqueue(rtlPeek.left);
        }
        return ltr.Count == 0 && rtl.Count == 0;
    }
}