[LeetCode] 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.

二叉树的最小深度。题目即是题意。我这里提供两种做法。

第一种思路是后序遍历。如果左孩子和右孩子有一个为NULL则最小深度为left + right + 1；否则就正常看两边较小的深度 + 1。

时间O(n)

空间O(n) - worse case，如果树不平衡，比如node全是左子树或者右子树；平均空间是O(log(n))

Java实现

class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = minDepth(root.left);
        int right = minDepth(root.right);
        if (left == 0 || right == 0) {
            return left + right + 1;
        }
        return Math.min(left, right) + 1;
    }
}

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function (root) {
    if (root == null) {
        return 0;
    }
    let left = minDepth(root.left);
    let right = minDepth(root.right);
    if (left == 0 || right == 0) {
        return left + right + 1;
    }
    return Math.min(left, right) + 1;
};

第二种思路是BFS。按照BFS的经典模板去遍历每一层的node。每弹出一个node的时候，判断当前node是否有左孩子和右孩子，如果没有，则说明当前这个node的深度应该是最小的了。

时间O(n)

空间O(n)

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        // corner case
        if (root == null) {
            return 0;
        }

        // normal case
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int depth = 1;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (cur.left == null && cur.right == null) {
                    return depth;
                }
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            depth++;
        }
        return depth;
    }
}

LeetCode 题目总结