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
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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
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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 ;
}
}
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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 ;
}
};
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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
}
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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
}
}