技術標籤：二叉樹leetcode二叉樹

DFS解法，遞迴左右，分別求左右子樹深度，然後求最小值。但是一開始沒考慮單邊的情況，直接返回min(left,right)。

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

BFS
1、通過雙佇列來實現

class Solution {
public:
    int minDepth(TreeNode* root) {
        vector<queue<TreeNode*> > v;
        queue<TreeNode*> q1;
        queue<TreeNode*> q2;
        v.push_back(q1);
        v.push_back(q2);
 

        int i=0;
        bool flag=true;
        TreeNode *pmove=root;
        v[i].push(pmove);
        while(flag&&pmove&&!v[i%2].empty())
        {
            while(flag&&!v[i%2].empty())
            {
            pmove=v[i%2].front();
            v[i%2].pop();
            if
 
(pmove->left)
            v[(i+1)%2].push(pmove->left);
            if(pmove->right)
            v[(i+1)%2].push(pmove->right);
            if(!pmove->left&&!pmove->right)
            flag=false;
            }
            ++i;
        }
        return i;
    }
};

繼續學習，發現自己還是太菜，不能靈活運用資料結構

class Solution {
public:
    int minDepth(TreeNode* root) {
        queue<pair<TreeNode*,int> > q;
        TreeNode *pmove=root;
        int dep=0;
        q.emplace(pmove,1);
        while(pmove&&!q.empty())
        {
            pmove=q.front().first;
            dep=q.front().second;
            q.pop();
            if(pmove->left)
            q.emplace(pmove->left,dep+1);
            if(pmove->right)
            q.emplace(pmove->right,dep+1);
            if(!pmove->right&&!pmove->left)
                break;
        }
        return dep;
    }
};

繼續優化

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