[LeetCode] 111. Minimum Depth of Binary Tree 二叉樹的最小深度

 

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.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its minimum depth = 2.

 

二叉樹的經典問題之最小深度問題就是就最短路徑的節點個數，還是用深度優先搜索 DFS 來完成，萬能的遞歸啊。首先判空，若當前結點不存在，直接返回0。然后看若左子結點不存在，那么對右子結點調用遞歸函數，並加1返回。反之，若右子結點不存在，那么對左子結點調用遞歸函數，並加1返回。若左右子結點都存在，則分別對左右子結點調用遞歸函數，將二者中的較小值加1返回即可，參見代碼如下：

 

解法一：

class Solution {
public:
    int minDepth(TreeNode* root) {
        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));
    }
};

 

我們也可以是迭代來做，層序遍歷，記錄遍歷的層數，一旦遍歷到第一個葉結點，就將當前層數返回，即為二叉樹的最小深度，參見代碼如下：

 

解法二：

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

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/111

 

類似題目：

Binary Tree Level Order Traversal

Maximum Depth of Binary Tree

 

參考資料：

https://leetcode.com/problems/minimum-depth-of-binary-tree/

https://leetcode.com/problems/minimum-depth-of-binary-tree/discuss/36153/My-concise-c%2B%2B-solution

https://leetcode.com/problems/minimum-depth-of-binary-tree/discuss/36071/BFS-C%2B%2B-8ms-Beats-99.94-submissions

 

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