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(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         LinkedList<TreeNode> que = new LinkedList<TreeNode>();
18         que.add(root);
19         int curCount = 1;
20         int nextCount = 0;
21         int curLevelMax = Integer.MIN_VALUE;
22         while(!que.isEmpty()){
23             TreeNode cur = que.poll();
24             curCount--;
25             curLevelMax = Math.max(curLevelMax, cur.val);
26             
27             if(cur.left != null){
28                 que.add(cur.left);
29                 nextCount++;
30             }
31             
32             if(cur.right != null){
33                 que.add(cur.right);
34                 nextCount++;
35             }
36             
37             if(curCount == 0){
38                 curCount = nextCount;
39                 nextCount = 0;
40                 res.add(curLevelMax);
41                 curLevelMax = Integer.MIN_VALUE;
42             }
43         }
44         return res;
45     }
46 }

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 }