题解 | #二叉树的深度#

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
  public:
    int TreeDepth(TreeNode* pRoot) {
      int res = 0;
      int tmp = 0;
      
      curision(pRoot, res, tmp);
      
      return res;
    }
  private:
    void curision(TreeNode *root, int &res, int &tmp) {
      if (root == nullptr) {
        res = std::max(res, tmp);
        return ;
      }
      
      ++tmp;
      curision(root->left, res, tmp);
      curision(root->right, res, tmp);
      //  回溯
      --tmp;
    }
};

简洁

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    int TreeDepth(TreeNode* pRoot) {
      if (pRoot == nullptr) {
        return 0;
      }
      
      //  1是当前节点，加上左右子树中最大的那部分
      return 1 + std::max(TreeDepth(pRoot->left), TreeDepth(pRoot->right));
    }
};