题目描述 
给定一棵二叉树的根节点 root ,请找出该二叉树中每一层的最大值。
 
示例1: 
输入:  root = [1,3,2,5,3,null,9]
输出:  [1,3,9]
解释: 
          1
         / \
        3   2
       / \   \  
      5   3   9 
 
示例2: 
输入:  root = [1,2,3]
输出:  [1,3]
解释: 
          1
         / \
        2   3
 
示例3: 
输入:  root = [1]
输出:  [1]
 
示例4: 
输入:  root = [1,null,2]
输出:  [1,2]
解释:       
           1 
            \
             2     
 
示例5: 
输入:  root = []
输出:  []
 
 
提示: 
    二叉树的节点个数的范围是 [0,104 ] 
    -231  <= Node.val <= 231  - 1 
 
 
 注意:本题与主站 515 题相同: https://leetcode.cn/problems/find-largest-value-in-each-tree-row/ 
解法 
方法一 
Python3 Java C++ Go Swift 
 1 
 2 
 3 
 4 
 5 
 6 
 7 
 8 
 9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 # Definition for a binary tree node. 
# class TreeNode: 
#     def __init__(self, val=0, left=None, right=None): 
#         self.val = val 
#         self.left = left 
#         self.right = right 
class   Solution : 
    def   largestValues ( self ,  root :  TreeNode )  ->  List [ int ]: 
        if  root  is  None : 
            return  [] 
        q  =  deque ([ root ]) 
        ans  =  [] 
        while  q : 
            t  =  - inf 
            for  _  in  range ( len ( q )): 
                node  =  q . popleft () 
                t  =  max ( t ,  node . val ) 
                if  node . left : 
                    q . append ( node . left ) 
                if  node . right : 
                    q . append ( node . right ) 
            ans . append ( t ) 
        return  ans 
 
 
 1 
 2 
 3 
 4 
 5 
 6 
 7 
 8 
 9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 /** 
 * Definition for a binary tree node. 
 * public class TreeNode { 
 *     int val; 
 *     TreeNode left; 
 *     TreeNode right; 
 *     TreeNode() {} 
 *     TreeNode(int val) { this.val = val; } 
 *     TreeNode(int val, TreeNode left, TreeNode right) { 
 *         this.val = val; 
 *         this.left = left; 
 *         this.right = right; 
 *     } 
 * } 
 */ 
class  Solution   { 
     public   List < Integer >   largestValues ( TreeNode   root )   { 
         List < Integer >   ans   =   new   ArrayList <> (); 
         if   ( root   ==   null )   { 
             return   ans ; 
         } 
         Deque < TreeNode >   q   =   new   ArrayDeque <> (); 
         q . offer ( root ); 
         while   ( ! q . isEmpty ())   { 
             int   t   =   Integer . MIN_VALUE ; 
             for   ( int   i   =   q . size ();   i   >   0 ;   -- i )   { 
                 TreeNode   node   =   q . poll (); 
                 t   =   Math . max ( t ,   node . val ); 
                 if   ( node . left   !=   null )   { 
                     q . offer ( node . left ); 
                 } 
                 if   ( node . right   !=   null )   { 
                     q . offer ( node . right ); 
                 } 
             } 
             ans . add ( t ); 
         } 
         return   ans ; 
     } 
} 
 
 
 1 
 2 
 3 
 4 
 5 
 6 
 7 
 8 
 9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 /** 
 * Definition for a binary tree node. 
 * struct TreeNode { 
 *     int val; 
 *     TreeNode *left; 
 *     TreeNode *right; 
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {} 
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 
 * }; 
 */ 
class   Solution   { 
public : 
     vector < int >   largestValues ( TreeNode *   root )   { 
         if   ( ! root )   return   {}; 
         queue < TreeNode *>   q {{ root }}; 
         vector < int >   ans ; 
         while   ( ! q . empty ())   { 
             int   t   =   INT_MIN ; 
             for   ( int   i   =   q . size ();   i   >   0 ;   -- i )   { 
                 auto   node   =   q . front (); 
                 q . pop (); 
                 t   =   max ( t ,   node -> val ); 
                 if   ( node -> left )   q . push ( node -> left ); 
                 if   ( node -> right )   q . push ( node -> right ); 
             } 
             ans . push_back ( t ); 
         } 
         return   ans ; 
     } 
}; 
 
 
 1 
 2 
 3 
 4 
 5 
 6 
 7 
 8 
 9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 /** 
 * Definition for a binary tree node. 
 * type TreeNode struct { 
 *     Val int 
 *     Left *TreeNode 
 *     Right *TreeNode 
 * } 
 */ 
func   largestValues ( root   * TreeNode )   [] int   { 
     var   ans   [] int 
     if   root   ==   nil   { 
         return   ans 
     } 
     var   q   =   [] * TreeNode { root } 
     for   len ( q )   >   0   { 
         t   :=   math . MinInt32 
         for   i   :=   len ( q );   i   >   0 ;   i --   { 
             node   :=   q [ 0 ] 
             q   =   q [ 1 :] 
             t   =   max ( t ,   node . Val ) 
             if   node . Left   !=   nil   { 
                 q   =   append ( q ,   node . Left ) 
             } 
             if   node . Right   !=   nil   { 
                 q   =   append ( q ,   node . Right ) 
             } 
         } 
         ans   =   append ( ans ,   t ) 
     } 
     return   ans 
} 
 
 
 1 
 2 
 3 
 4 
 5 
 6 
 7 
 8 
 9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
40 
41 
42 
43 
44 
45 
46 
47 /* class TreeNode { 
*     var val: Int 
*     var left: TreeNode? 
*     var right: TreeNode? 
*     init() { 
*         self.val = 0 
*         self.left = nil 
*         self.right = nil 
*     } 
*     init(_ val: Int) { 
*         self.val = val 
*         self.left = nil 
*         self.right = nil 
*     } 
*     init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) { 
*         self.val = val 
*         self.left = left 
*         self.right = right 
*     } 
* } 
*/ 
class   Solution   { 
     func   largestValues ( _   root :   TreeNode ?)   ->   [ Int ]   { 
         var   ans   =   [ Int ]() 
         guard   let   root   =   root   else   { 
             return   ans 
         } 
         var   q   =   [ TreeNode ]() 
         q . append ( root ) 
         while   ! q . isEmpty   { 
             var   t   =   Int . min 
             for   _   in   0. .< q . count   { 
                 let   node   =   q . removeFirst () 
                 t   =   max ( t ,   node . val ) 
                 if   let   left   =   node . left   { 
                     q . append ( left ) 
                 } 
                 if   let   right   =   node . right   { 
                     q . append ( right ) 
                 } 
             } 
             ans . append ( t ) 
         } 
         return   ans 
     } 
}