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:
java
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;
}
}
java
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 。