Comments (2)
解题思路
- 深度优先算法, 递归找到层数最深的节点.
代码
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() { }
TreeNode(int val) { this.val = val; }
}
public static int func(TreeNode root) {
if (null == root) return 0;
return Math.max(func(root.left) + 1, func(root.right) + 1);
}
from pisecespeng.record.me.
LeetCode题解
解题思路
- 广度优先算法,
- 将每层的节点暂时存放在队列中,
- 且不断推出该节点与推入该节点的子节点们(如果存在的话),
- 以上循环, 记录循环次数.
代码
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() { }
TreeNode(int val) { this.val = val; }
}
public static int func(TreeNode root) {
if (root == null) {
return 0;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
int ans = 0;
while (!queue.isEmpty()) {
int size = queue.size();
while (size > 0) {
TreeNode node = queue.poll();
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
size--;
}
ans++;
}
return ans;
}
from pisecespeng.record.me.
Related Issues (20)
- 两数相加 HOT 2
- 买卖股票的最佳时机 HOT 3
- 买卖股票的最佳时期2 HOT 3
- 二叉树的层序遍历 HOT 2
- 二进制求和 HOT 4
- 从尾到头打印链表 HOT 2
- 位1的个数 HOT 4
- 删除排序数组中的重复项 HOT 2
- 删除链表的倒数第N个节点 HOT 3
- 加一 HOT 2
- 反转字符串 HOT 2
- 反转链表 HOT 3
- 只出现一次的数字 HOT 2
- 合并两个有序数组 HOT 3
- 合并两个有序链表 HOT 2
- 回文数 HOT 2
- 回文链表 HOT 3
- 复制带随机指针的链表 HOT 2
- 外观数列 HOT 3
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