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.

给一个二叉树，找出它的最小深度。最小深度是从根节点向下到最近的叶节点的最短路径，就是最短路径的节点个数。

解法1：DFS

解法2: BFS

Java: DFS, Time Complexity: O(n)， Space Complexity: O(n)

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null) {
return 1 + minDepth(root.right);
} else if (root.right == null) {
return 1 + minDepth(root.left);
} else {
return 1 + Math.min(minDepth(root.left), minDepth(root.right));
}
}
}　

Java: BFS

public class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0;
int depth = 1;

while (!q.isEmpty()) {
TreeNode node = q.poll();
curLevel--;
if (node.left == null && node.right == null) {
return depth;
}
if (node.left != null) {
q.offer(node.left);
nextLevel++;
}
if (node.right != null) {
q.offer(node.right);
nextLevel++;
}
if (curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
depth++;
}
}
return depth;
}
}　　

Python:

class Solution(object):
def minDepth(root):
if root is None:
return 0

# Base Case : Leaf node.This acoounts for height = 1
if root.left is None and root.right is None:
return 1

if root.left is None:
return minDepth(root.right) + 1

if root.right is None:
return minDepth(root.left) + 1

return min(minDepth(root.left), minDepth(root.right)) + 1　　

Python:

class Solution:
# @param root, a tree node
# @return an integer
def minDepth(self, root):
if root is None:
return 0

if root.left and root.right:
return min(self.minDepth(root.left), self.minDepth(root.right)) + 1
else:
return max(self.minDepth(root.left), self.minDepth(root.right)) + 1　　

C++:

/**
* Definition for binary tree
* 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 == NULL) return 0;
if (root->left == NULL && root->right == NULL) return 1;

if (root->left == NULL) return minDepth(root->right) + 1;
else if (root->right == NULL) return minDepth(root->left) + 1;
else return 1 + min(minDepth(root->left), minDepth(root->right));
}

};

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[LeetCode] 104. Maximum Depth of Binary Tree 二叉树的最大深度

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