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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