后序遍历
class Solution {
public:
int dfs(TreeNode* node) {
if (node == nullptr) return 0;
if (node->left == nullptr && node->right != nullptr) {
return 1 + dfs(node->right);
}
if (node->right == nullptr && node->left != nullptr) {
return 1 + dfs(node->left);
}
return 1 + min(dfs(node->left), dfs(node->right));
}
int minDepth(TreeNode* root) {
return dfs(root);
}
};
标签:node,right,return,nullptr,dfs,111,二叉树,LeetCode,left
From: https://www.cnblogs.com/hjy94wo/p/16664255.html