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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.com/problems/find-largest-value-in-each-tree-row/

解法

“BFS 层次遍历”实现。

Python3

# 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:
            n = len(q)
            t = float('-inf')
            for _ in range(n):
                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

Java

/**
 * 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) {
        if (root == null) {
            return Collections.emptyList();
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        List<Integer> ans = new ArrayList<>();
        while (!q.isEmpty()) {
            int t = Integer.MIN_VALUE;
            for (int i = 0, n = q.size(); i < n; ++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;
    }
}

C++

/**
 * 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;
        vector<int> ans;
        q.push(root);
        while (!q.empty())
        {
            int t = INT_MIN;
            for (int i = 0, n = q.size(); i < n; ++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;
    }
};

Go

/**
 * 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 {
		n := len(q)
		t := math.MinInt32
		for i := 0; i < n; 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
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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