題目:
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.
解題思路1:設置一個變量存儲當前記錄下的最小深度,采用先序遍歷的方法訪問樹的每一個節點,設置一個變量表示當前節點所在的層次,如果一個節點沒有子節點,那麼就比較該節點的深度與當前的最小深度,選擇兩者之中較小的作為當前的最小深度。
代碼:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int minDepth(TreeNode *root) { int curr_depth=0,min_depth=1000000; return PreorderTraverse(root,&curr_depth,min_depth); } private: int PreorderTraverse(TreeNode *root, int *curr_depth, int min_depth){ if(root==nullptr)return 0; (*curr_depth)++; if(root->left!=nullptr){ min_depth=PreorderTraverse(root->left, curr_depth, min_depth); } if(root->right!=nullptr){ min_depth=PreorderTraverse(root->right, curr_depth, min_depth); } if((root->left==nullptr)&&(root->right==nullptr)){ min_depth=min(*curr_depth,min_depth); } (*curr_depth)--; return min_depth; } };
代碼2:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int minDepth(TreeNode *root) { return minDepth(root,false); } private: int minDepth(TreeNode *root, bool hasbrother){ if(!root){ return hasbrother==false?0:INT_MAX; } return 1+min(minDepth(root->left,root->right!=NULL),minDepth(root->right,root->left!=NULL)); } };