Question

剑指 Offer II 044. 二叉树每层的最大值

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

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}