题目描述
题目链接:111. Minimum Depth of Binary Tree
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.
例子
例子 1
Input:
[3,9,20,null,null,15,7]
Output: 2
例子 2
Input:
root = [2,null,3,null,4,null,5,null,6]
Output: 5
Note
- A leaf is a node with no children.
解题思路
这道题求二叉树的最浅深度,可以用 DFS 或者 BFS 来求解:
方法一: DFS
通过 DFS 遍历二叉树,并在遇到叶子结点时更新结果即可。为了加快速度,可以在当前深度达到当前最小深度时停止遍历该子树。 代码如下:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
#include <bits/stdc++.h>
class Solution {
public:
int minDepth(TreeNode* root) {
if (!root) return 0;
int min_depth = INT_MAX;
dfs(root, 0, min_depth);
return min_depth;
}
private:
void dfs(TreeNode* node, int current_depth, int& min_depth) {
if (!node || current_depth >= min_depth) {
return;
}
current_depth++;
if (!node->left && !node->right) {
min_depth = current_depth;
}
dfs(node->left, current_depth, min_depth);
dfs(node->right, current_depth, min_depth);
}
};
- 时间复杂度: O(n)
- 空间复杂度: O(h) – 遍历深度为最大二叉树最大深度
方法二: BFS
结合 std::queue
和 BFS 按层遍历二叉树,在遇到第一个叶子节点时即可返回当前深度作为结果,代码如下:
class Solution {
public:
int minDepth(TreeNode* root) {
// BFS Solution
queue<TreeNode*>q;
if(root==NULL)return 0;
q.push(root);
int level=0;
while(!q.empty())
{
int s=q.size();
level++;
while(s--)
{
auto cur=q.front();
q.pop();
if(cur->left)q.push(cur->left);
if(cur->right)q.push(cur->right);
if(!cur->left&&!cur->right)return level; // This must be leaf node so return it
}
}
return level;
}
};
- 时间复杂度: O(n)
- 空间复杂度: O(w)
GitHub 代码同步地址: 111.MinimumDepthOfBinaryTree.cpp
其他题目: GitHub: Leetcode-C++-Solution 博客: Leetcode-Solutions