leetcode111. 二叉树的最小深度

问题描述：

给定一个二叉树，找出其最小深度。

最小深度是从根节点到最近叶子节点的最短路径上的节点数量。

说明: 叶子节点是指没有子节点的节点。

示例:

给定二叉树 3,9,20,null,null,15,7,

3

/ \

9 20

/ \

15 7

返回它的最小深度 2.

思路：见代码

上代码，拿去即可运行：

package onlyqi.daydayupgo06.leetcode;
 
public class TreeNode {
 
    private Integer value;
    private TreeNode left;
    private TreeNode right;
 
 
    public TreeNode() {
    }
 
    public TreeNode(Integer value) {
        this.value=value;
    }
 
    public TreeNode(Integer value, TreeNode left, TreeNode right) {
        this.value = value;
        this.left = left;
        this.right = right;
    }
 
    public Integer getValue() {
        return value;
    }
 
    public void setValue(Integer value) {
        this.value = value;
    }
 
    public TreeNode getLeft() {
        return left;
    }
 
    public void setLeft(TreeNode left) {
        this.left = left;
    }
 
    public TreeNode getRight() {
        return right;
    }
 
    public void setRight(TreeNode right) {
        this.right = right;
    }
}

    public static void main(String[] args) {
        TreeNode treeNode1 = new TreeNode(1);
        TreeNode treeNode2 = new TreeNode(2);
        TreeNode treeNode3 = new TreeNode(3);
        TreeNode treeNode4 = new TreeNode(4);
        TreeNode treeNode5 = new TreeNode(5);
        TreeNode treeNode6 = new TreeNode(6);
        TreeNode treeNode7 = new TreeNode(7);
        treeNode2.setLeft(treeNode4);
        treeNode2.setRight(treeNode5);
        treeNode1.setLeft(treeNode2);
        treeNode1.setRight(treeNode3);
        treeNode5.setLeft(treeNode6);
//        treeNode3.setLeft(treeNode7);

        System.out.println(getHeightBFS1(treeNode1));

    public static int getHeightBFS1(TreeNode root) {
        if (root == null) return 0;

        int height = 0;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            height++;

            for (int i = 0; i < levelSize; i++) {

                TreeNode node = queue.poll();
                if (node.getLeft() == null && node.getRight() == null) {
                    return height;
                }
                if (node.getLeft() != null) queue.offer(node.getLeft());
                if (node.getRight() != null) queue.offer(node.getRight());
            }
        }
        return height;
    }
}

运行结果：

我要刷300道算法题，第142道 。 尽快刷到200，每天搞一道 。