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[Leetcode]111. Minimum Depth of Binary Tree(C++)

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:

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/**
* 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:

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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