二叉树的最小深度

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.

c++ 广度优先

/*** 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) {std::queue<TreeNode*> nodeQueue;int res = 0;if (root) {if (!root->left && !root->right) {return 1;} else {res++;nodeQueue.push(root);}} else {return 0;}while(!nodeQueue.empty()) {auto nodeQueueSize = nodeQueue.size();for (int i=0; i<nodeQueueSize; i++) {auto node = nodeQueue.front();if (!node->left && !node->right) {return res;} else {if (node->left) {nodeQueue.push(node->left);}if (node->right) {nodeQueue.push(node->right);}}nodeQueue.pop();}res++;}return res;}
};

c++ 递归

/*** 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;}if (!root->left && !root->right) {return 1;}auto min_depth = INT_MAX;if (root->left) {min_depth = min(minDepth(root->left), min_depth);}if (root->right) {min_depth = min(minDepth(root->right), min_depth);}return min_depth + 1;}
};

Python 递归

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def minDepth(self, root: TreeNode) -> int:if not root:return 0children = [root.left, root.right]if not any(children):return 1min_depth = float('inf')for c in children:if (c):min_depth = min(self.minDepth(c), min_depth)return min_depth + 1