题目:
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.
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.
解题:
深度优先搜索方式,分别找到左右子树最小+1;广度优先搜索,逐层遍历找到叶子节点返回层数。
实现:
- DFS 空间复杂度O(N) 时间复杂度 O(N)
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));
}
};
实现:
- BFS 空间复杂度O(N) 时间复杂度 O(N)
/**
* 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) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if (!root){
return 0;
}
int depth = 0;
queue<TreeNode *> q;
q.push(root);
while(!q.empty()){
depth++;
int level_size = q.size();
for(int i=0; i<level_size; i++){
TreeNode *node = q.front();
q.pop();
if (!node->left&&!node->right){
return depth;
}
if (node->left){
q.push(node->left);
}
if (node->right){
q.push(node->right);
}
}
}
return depth;
}
};
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