题解 | #二叉树的最小深度#
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.Queue;
import java.util.LinkedList;
public class Solution {
public int run(TreeNode root) {
if(root == null)
return 0;
if(root.left == null && root.right == null)
return 1;
int depth = 0; //记录层数遍历的深度
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while(!queue.isEmpty()){
int len = queue.size();
depth++;
//len表示树的每一层的节点数
for(int i = 0; i < len; i++){
TreeNode node = queue.poll();
if(node.left == null && node.right == null)
return depth;
if(node.left != null)
queue.offer(node.left);
if(node.right != null)
queue.offer(node.right);
}
}
return depth;
}
}
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.Queue;
import java.util.LinkedList;
public class Solution {
public int run(TreeNode root) {
if(root == null)
return 0;
if(root.left == null && root.right == null)
return 1;
int depth = 0; //记录层数遍历的深度
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while(!queue.isEmpty()){
int len = queue.size();
depth++;
//len表示树的每一层的节点数
for(int i = 0; i < len; i++){
TreeNode node = queue.poll();
if(node.left == null && node.right == null)
return depth;
if(node.left != null)
queue.offer(node.left);
if(node.right != null)
queue.offer(node.right);
}
}
return depth;
}
}