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.
Analyse: For the current node, minimum depth equals to 1 + min(depth of left subtree, depth of right subtree). We can do it recursively or iteratively.
1. Recursion
New Version: 12ms
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int minDepth(TreeNode* root) { 13 if(!root) return 0; 14 if(!root->left && !root->right) return 1; 15 16 if(!root->left) return 1 + minDepth(root->right); 17 if(!root->right) return 1 + minDepth(root->left); 18 return 1 + min(minDepth(root->left), minDepth(root->right)); 19 } 20 };
Runtime: 12ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int minDepth(TreeNode* root) { 13 if(!root) return 0; 14 int leftDepth = minDepth(root->left); 15 int rightDepth = minDepth(root->right); 16 if(leftDepth == 0 && rightDepth == 0) return 1; 17 if(leftDepth == 0) leftDepth = INT_MAX; 18 if(rightDepth == 0) rightDepth = INT_MAX; 19 20 return min(leftDepth, rightDepth) + 1; 21 } 22 };
2. Iteration: BFS. When it comes to a leaf node, then return its current depth.
Runtime: 12ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int minDepth(TreeNode* root) { 13 if(!root) return 0; 14 15 queue<TreeNode* > qu; 16 qu.push(root); 17 int level = 1; 18 while(!qu.empty()){ 19 int n = qu.size(); 20 while(n--){ 21 TreeNode* current = qu.front(); 22 qu.pop(); 23 if(!current->left && !current->right) return level; 24 if(current->left) qu.push(current->left); 25 if(current->right) qu.push(current->right); 26 } 27 level++; 28 } 29 return level; 30 } 31 };