Minimum Depth of Binary Tree

Problem

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 1:

Input: root = [3,9,20,null,null,15,7]
Output: 2

Example 2:

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

 

Constraints:

• The number of nodes in the tree is in the range [0, 105].
• -1000 <= Node.val <= 1000

Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function(root) {
    if (!root) {
        return 0;
    } else if (root.left && root.right) {
        return 1 + Math.min(minDepth(root.left), minDepth(root.right));
    } else {
        return 1 + minDepth(root.left ?? root.right);
    }
};

We use standard DFS to find the minimum depth of the tree root.

Optimization

A far better and more optimized solution is to use BFS, since DFS needs to traverse every branch of the tree before terminating (BFS does not).

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function(root) {
    let depth = 0;
    const queue = root ? [root] : [];

    while (queue.length) {
        depth++;
        const n = queue.length;
        for (let i = 0; i < n; i++) {
            const node = queue.shift();

            if (!node.left && !node.right) { // leaf node
                return depth;
            }
            if (node.left) {
                queue.push(node.left);
            }
            if (node.right) {
                queue.push(node.right);
            }
        }
    }
    return depth;
};

We do a level order traversal using BFS, and return the current depth when a leaf node is found.