Minimum Depth of Binary Tree ——LeetCode

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.

https://leetcode.com/problems/minimum-depth-of-binary-tree/

题意就是给定一个二叉树，找出从根节点到叶子节点最低的高度，直接写了个递归，算出所有叶子节点的高度，取最小值。

另外一种很显然就用层次遍历（BFS）， 就是按层次遍历这棵树，发现当前节点是叶子节点，就直接返回这个节点的高度。这里有个问题就是怎么记录当前的高度呢？我是这么做的：设置三个变量，upRow，downRow，height，分别代表队列中上一行节点数，下一行节点数，高度，当队列中上一行节点数为0，那么高度+1，把下一行的节点数量（downRow）赋给上一行（upRow），循环不变式是队列非空（!queue.isEmpty()）。

Talk is cheap。

import java.util.LinkedList;

import java.util.List;



/**

 * Created with IntelliJ IDEA.

 * User: Blank

 * Date: 2015/3/21

 */

public class MinimumDepthofBinaryTree {



    public int minDepth(TreeNode root) {

        int left = Integer.MAX_VALUE, right = Integer.MAX_VALUE;

        if (root == null)

            return 0;

        if (root.right == null && root.left == null)

            return 1;

        if (root.left != null)

            left = minDepth(root.left) + 1;

        if (root.right != null)

            right = minDepth(root.right) + 1;

        return Math.min(left, right);

    }



    public int minDepth2(TreeNode root) {

        if (root == null) {

            return 0;

        }

        int upRow = 1, downRow = 0, height = 0;

        List<TreeNode> queue = new LinkedList<>();

        queue.add(root);

        while (!queue.isEmpty()) {

            TreeNode node = queue.get(0);

            queue.remove(0);

            if (node.left == null && node.right == null)

                return height + 1;

            if (node.left != null) {

                queue.add(node.left);

                downRow++;

            }

            if (node.right != null) {

                queue.add(node.right);

                downRow++;

            }

            upRow--;

            if (upRow == 0) {

                height++;

                upRow = downRow;

                downRow = 0;

            }

        }

        return height;

    }

}