LeetCode 515. Find Largest Value in Each Tree Row

原题链接在这里：https://leetcode.com/problems/find-largest-value-in-each-tree-row/description/

题目：

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]

题解：

可以采用BFS, 类似Binary Tree Level Order Traversal. 每层算最大值加入res中.

Time Complexity: O(n).

Space: O(n).

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public List<Integer> largestValues(TreeNode root) {
18         List<Integer> res = new ArrayList<>();
19         if(root == null){
20             return res;
21         }
22         
23         LinkedList<TreeNode> que = new LinkedList<>();
24         que.add(root);
25         while(!que.isEmpty()){
26             int size = que.size();
27             int max = Integer.MIN_VALUE;
28             while(size-- > 0){
29                 TreeNode cur = que.poll();
30                 max = Math.max(max, cur.val);
31                 if(cur.left != null){
32                     que.add(cur.left);
33                 }
34                 
35                 if(cur.right != null){
36                     que.add(cur.right);
37                 }
38             }
39             
40             res.add(max);
41         }
42         
43         return res;
44     }
45 }

也可以DFS. 用depth来标记res中的index位置.

Time Complexity: O(n).

Space: O(logn). stack space.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public List<Integer> largestValues(TreeNode root) {
12         List<Integer> res = new ArrayList<Integer>();
13         if(root == null){
14             return res;
15         }
16         
17         dfs(root, res, 0);
18         return res;
19     }
20     
21     private void dfs(TreeNode root, List<Integer> res, int depth){
22         if(root == null){
23             return;
24         }
25         
26         if(depth == res.size()){
27             // 之前没有的碰到的深度.
28             res.add(root.val);
29         }else{
30             // 之前有平级的深度.
31             res.set(depth, Math.max(res.get(depth), root.val));
32         }
33         
34         dfs(root.left, res, depth+1);
35         dfs(root.right, res, depth+1);
36     }
37 }

 

posted @ 2017-11-27 13:09  Dylan_Java_NYC  阅读(308)  评论(0)  编辑  收藏  举报