Leetcode 866. Smallest Subtree with all the Deepest Nodes

https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/description/

好想好写的写法就是两次遍历

class Solution {
public:
	
	int maxDep(TreeNode* root) {
		if (!root) return 0;
		return max(maxDep(root->left), maxDep(root->right)) + 1;
	}
	
	TreeNode* find(TreeNode* root,  int cur_dep, int max_dep) {
		if(!root) return NULL;
		
		TreeNode* left = find(root->left,  cur_dep+1, max_dep);
		TreeNode* right = find(root->right, cur_dep+1, max_dep);

		if (cur_dep == max_dep) {
			
			return root;
		}
		
		if (left && right) return root;
		if (left) return left;
		if (right) return right;
		return NULL;
	}
	
	
	TreeNode* subtreeWithAllDeepest(TreeNode* root) {
		if (!root) return root;
		int max_dep = maxDep(root);
		return find(root, 1, max_dep);
	}
};

另外后续遍历的时候，子树深度 还有距离根节点的深度是可以同时获得的，于是能拿到总的深度，

https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/discuss/146821/One-Traversal-C++-Simplest-one!!

这个解有意思

int utils(TreeNode *root, int c_d, int &max_d, TreeNode* &res) {
    if(root == nullptr)
        return 0;
    int l = utils(root->left, c_d + 1, max_d, res);
    int r = utils(root->right, c_d + 1, max_d, res);
    int h = max(l, r);
    
    if(l == r && c_d + h >= max_d) {
        max_d = c_d + h;
        res = root;
    }

    return h + 1;
}
TreeNode* subtreeWithAllDeepest(TreeNode* root) {
    TreeNode *res = root;
    int c_d = 0, max_d = INT_MIN;
    utils(root, c_d, max_d, res);
    return res;
}