[日常刷题]LeetCode第十一天

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101. 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.

Solution in C++：

• 关键点：递归中的处理部分和昨天的相同树差不多
• 思路：总体思路和相同树差不多，就是比较的子树的位置不一样，交换一下就可以了。

bool isSymme(TreeNode* left, TreeNode* right){
        
        if (left == nullptr && right == nullptr)
            return true;
        else if(left == nullptr || right == nullptr)
            return false;
        
        if(left->val != right->val)
            return false;
        
        bool left_left = isSymme(left->left, right->right);
        bool right_right = isSymme(left->right, right->left);
        return left_left && right_right;    
    }
    
    bool isSymmetric(TreeNode* root) {
        if (root == nullptr)
            return true;
        else 
            return isSymme(root->left, root->right);
    }

---

104. Maximum Depth of Binary Tree

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its depth = 3.

Solution in C++：

• 关键点：左右子树不可重复加
• 思路：开始打算用层次遍历的方法去做，但是后来发现好像递归就可以解决，每次扫描一层就加一。

int maxDepth(TreeNode* root) {
        
        if (root == nullptr)
            return 0;
        else{
            
            int llenth = maxDepth(root->left);
            
            int rlenth = maxDepth(root->right);
            
            if(rlenth >= llenth)
                return rlenth + 1;
            else
                return llenth + 1;
        }
    }

---

小结

今天主要两个都是对于树的遍历问题的应用。一个是通过定义不同规则来判断结构问题，另一个是有点层次扫描的感觉。

• 树的遍历