LeetCode – Minimum Depth of Binary Tree

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;
}
};
``````