Leetcode | Minimum/Maximum Depth of Binary Tree

Minimum Depth of Binary Tree

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

BFS碰到一个叶子结点就可以了。

 1 class Solution {

 2 public:

 3     int minDepth(TreeNode *root) {

 4         if (root == NULL) return NULL;

 5         

 6         queue<TreeNode*> q;

 7         q.push(root);

 8         q.push(NULL);

 9         int h = 1;

10         while (q.size() > 1) {

11             TreeNode* p = q.front();

12             q.pop();

13             

14             if (p == NULL) { h++; q.push(NULL); continue;}

15             if (p->left == NULL && p->right == NULL) break;

16             if (p->left) q.push(p->left);

17             if (p->right) q.push(p->right);

18         }

19         return h;

20     }

21     

22 };

这里用个NULL指针作哨兵，作为层的结束标志。所有遍历完时，q.size() == 1（q里面只有NULL一个点)。 不过这里因为只要到达叶子结点就会退出，所以不存在死循环的问题。

第三次，bugfree一遍通过。

 1 class Solution {

 2 public:

 3     int minDepth(TreeNode *root) {

 4         if (root == NULL) return 0;

 5         vector<vector<TreeNode*> > layers(2);

 6         int cur = 0, next = 1, layer = 1;

 7         layers[cur].push_back(root);

 8         while (!layers[cur].empty()) {

 9             layers[next].clear();

10             for (auto node: layers[cur]) {

11                 if (node->left == NULL && node->right == NULL) return layer;

12                 if (node->left) layers[next].push_back(node->left);

13                 if (node->right) layers[next].push_back(node->right);

14             }

15             cur = !cur, next = !next;

16             layer++;

17         }

18         return layer;

19     }

20 };

 Maximum Depth of Binary Tree

同样是用bfs好记录层数，然后bfs结束返回值就行了。 

 1 class Solution {

 2 public:

 3     int maxDepth(TreeNode *root) {

 4         if (root == NULL) return 0;

 5         vector<vector<TreeNode*> > layers(2);

 6         int cur = 0, next = 1, layer = 0;

 7         layers[cur].push_back(root);

 8         while (!layers[cur].empty()) {

 9             layers[next].clear();

10             for (auto node: layers[cur]) {

11                 if (node->left) layers[next].push_back(node->left);

12                 if (node->right) layers[next].push_back(node->right);

13             }

14             cur = !cur, next = !next;

15             layer++;

16         }

17         return layer;

18     }

19 };