Skip to content

Latest commit

 

History

History
21 lines (20 loc) · 955 Bytes

111. Minimum Depth of Binary Tree.md

File metadata and controls

21 lines (20 loc) · 955 Bytes

思路

求二叉树的最小深度。依次分为以下情况:

  • 若为空树,则返回0;
  • 否则,若左右子树都为空,返回1;
  • 否则,若左子树为空,那么最小深度就为右子树的最小深度加1;若右子树为空,那么最小深度就为左子树的最小深度加1;
  • 否则(即左右子树都不空),那么最小深度就是min(1+左子树最小深度, 1+右子树最小深度);

C++

class Solution {
public:
    int minDepth(TreeNode* root) {
        if(root == NULL) return 0;
        else if(root -> left == NULL && root -> right == NULL) return 1;
        else if(root -> left == NULL) return 1 + minDepth(root -> right);
        else if(root -> right == NULL) return 1 + minDepth(root -> left);
        return 1 + min(minDepth(root -> left), minDepth(root -> right));
    }
};