111 Minimum Depth of Binary Tree 二叉树的最小深度
Description:
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.
题目描述:
给定一个二叉树,找出其最小深度。
最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
说明: 叶子节点是指没有子节点的节点。
示例:
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回它的最小深度 2.
思路:
参考LeetCode #104 Maximum Depth of Binary Tree 二叉树的最大深度
跟最大深度略有不同, 找到叶结点即可返回叶结点所在层数, 注意根结点不是叶结点, 所以[1, 2]应当返回 2, 而不是1.
- 递归
- 迭代
时间复杂度O(n), 空间复杂度O(n), n为树中结点数
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
public:
int minDepth(TreeNode* root)
{
if (!root) return 0;
queue<TreeNode*> q;
q.push(root);
int result = 0;
while (!q.empty())
{
int nums = q.size();
result++;
while (nums)
{
TreeNode* cur = q.front();
q.pop();
if (!cur -> left and !cur -> right) return result;
if (cur -> left) q.push(cur -> left);
if (cur -> right) q.push(cur -> right);
nums--;
}
}
return result;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if (root == null) return 0;
int left = minDepth(root.left);
int right = minDepth(root.right);
if (root.left == null || root.right == null) return 1 + left + right;
return Math.min(left, right) + 1;
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def minDepth(self, root: TreeNode) -> int:
if not root:
return 0
left = self.minDepth(root.left)
right = self.minDepth(root.right)
if not left or not right:
return 1 + left + right
return min(left, right) + 1