111. Minimum Depth of Binary Tree

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.

Solution

Approach 1: BFS iteration.

Approach 2: recursive -- but be careful when a node has only 1 child.

Code (Python)

Approach 1:

    def minDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # BFS
        if not root:
            return 0
        level = [root]
        depth = 1
        while level:
            new_level = []
            for node in level:
                if node.left:
                    new_level.append(node.left)
                if node.right:
                    new_level.append(node.right)
                if not node.left and not node.right:
                    return depth
            if new_level:
                level = new_level
                depth += 1
        return depth

Approach 2:

    def minDepth(self, root):
        # recursive
        # this is bad, because it won't count to the very bottom for nodes that has only 1 child
        #if not root:
        #    return 0
        #return min(self.minDepth(root.left), self.minDepth(root.right)) + 1        
        if not root:
            return 0
        if not root.left and not root.right:
            return 1
        if not root.left or not root.right:
            return self.minDepth(root.left) + 1 if root.left else self.minDepth(root.right) + 1
        return min(self.minDepth(root.left), self.minDepth(root.right)) + 1

Code (C++)

Approach 1:

/**
 * 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 == NULL) return 0;
        queue<TreeNode*> nodeQueue;
        nodeQueue.push(root);
        int level = 0;
        while (!nodeQueue.empty())
        {
            level++;
            int nodeQueueSize = nodeQueue.size();
            for (int i = 0; i < nodeQueueSize; ++i) // BFS with each level separated.
            {
                TreeNode *node = nodeQueue.front();
                nodeQueue.pop();
                if (node->left == NULL && node->right == NULL)
                    return level;
                if (node->left) nodeQueue.push(node->left);
                if (node->right) nodeQueue.push(node->right);
            }
        }
        return level;
    }
};