515. Find Largest Value in Each Tree Row查找一行中的最大值

[抄题]:

You need to find the largest value in each row of a binary tree.

Example:

Input: 

          1
/ \
3 2
/ \ \
5 3 9 Output: [1, 3, 9]

[暴力解法]:

时间分析:

空间分析:

[优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

[思维问题]:

不知道怎么确定每一行的大小:不熟悉bfs。其中q每次只存了一行,所以size就是当前数组的大小

[英文数据结构或算法,为什么不用别的数据结构或算法]:

Queue<TreeNode> q = new LinkedList<>(); 因为都可以随便动?

[一句话思路]:

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[五分钟肉眼debug的结果]:

[总结]:

BFS要点:(3先生)先加头、先判Empty、先取长度

[复杂度]:Time complexity: O(n) Space complexity: O(n)

图是v+e 树就是点数n

[算法思想:迭代/递归/分治/贪心]:

[关键模板化代码]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

[代码风格] :

[是否头一次写此类driver funcion的代码] :

// package whatever; // don't place package name!

import java.io.*;
import java.util.*;
import java.lang.*; class TreeNode
{
int val;
TreeNode left, right; //parameter is another item
TreeNode(int item) {
val = item;
left = right = null;
}
} class Solution {
TreeNode root; public List<Integer> largestValues(TreeNode root) {
//ini: q, int max, Array
int max = Integer.MIN_VALUE;
//implement by linkedlist
Queue<TreeNode> q = new LinkedList<>();
List<Integer> answer = new ArrayList<Integer>(); //cc
if (root == null) return answer; q.offer(root);
while (!q.isEmpty()) {
//
int size = q.size(); for (int i = 0; i < size; i++) {
TreeNode cur = q.poll();
max = Math.max(cur.val, max);
if (cur.left != null) q.offer(cur.left);
if (cur.right != null) q.offer(cur.right);
}
//add to answer
answer.add(max);
//renew max
max = Integer.MIN_VALUE;
} return answer;
}
} class MyCode {
public static void main (String[] args) {
Solution tree = new Solution();
tree.root = new TreeNode(1);
tree.root.left = new TreeNode(2);
tree.root.right = new TreeNode(3);
tree.root.left.left = new TreeNode(4);
tree.root.left.right = new TreeNode(5); //TreeNode t = tree.upsideDownBinaryTree(tree.root);
List<Integer> answer = tree.largestValues(tree.root);
int size = answer.size();
for (int i = 0; i < size; i++)
System.out.println("answer[i] = " + answer.get(i));
}
}

[潜台词] :

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