111. Minimum Depth of Binary Tree

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problem

solution

dfs

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class Solution {
public:
    int minDepth(TreeNode* root) {
        // dfs
        if(!root) return 0;
        if(!root->left) return 1+minDepth(root->right);
        if(!root->right) return 1+minDepth(root->left);
        return 1+min(minDepth(root->left), minDepth(root->right));
    }
};

bfs

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class Solution {
public:
    int minDepth(TreeNode* root) {
        // bfs
        if(!root) return 0;
        queue<TreeNode*> q;
        q.push(root);
        int minDepth = 1;
        while(!q.empty()){
            int size  = q.size();
            for(int i=0;i<size; ++i){
                TreeNode *p = q.front();
                q.pop();
                
                if(!p->left && !p->right) return minDepth;
                else if(p->left && p->right){
                    q.push(p->left);
                    q.push(p->right);
                }
                else if(!p->left){
                    q.push(p->right);
                }
                else if(!p->right){
                    q.push(p->left);
                }
                
            }
            minDepth++;
        }
        return minDepth;
    }
};

analysis

  • backtracking
    time complexity O(n)
    space complexity O(logN)
  • bfs
    time complexity O(n)
    space complexity O(n) , N/2 樹最底層的節點數量