101.Symmetric Tree

Tags: Easy Depth-first Search Bread-first Search

Links: https://leetcode.com/problems/symmetric-tree/

---

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.

---

/**
 * 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) {
        return root ? isSymmetric(root -> left, root -> right) : true;
    }

    bool isSymmetric(TreeNode *left, TreeNode *right) {
        if (!left && !right) return true;
        if (!left || !right) return false;

        return left -> val == right -> val 
            && isSymmetric(left -> left, right -> right) 
            && isSymmetric(right -> left, left -> right);
    }
};

非递归写法：

/**
 * 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) return true;

        stack<TreeNode *> s;
        s.push(root -> left); 
        s.push(root -> right);
        while (!s.empty()) {
            TreeNode *p = s.top();
            s.pop();
            TreeNode *q = s.top();
            s.pop();

            if (!p && !q) continue;
            if (!p || !q) 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;
    }
};

这道题主要是注意考虑的是对称，所以递归写法检验的是left -> left, right -> right和left -> right, right -> left。