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515. 在每个树行中找最大值

English Version

题目描述

您需要在二叉树的每一行中找到最大的值。

示例：

输入: 

          1
         / \
        3   2
       / \   \  
      5   3   9 

输出: [1, 3, 9]

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 = collections.deque([root])
        res = []
        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)
            res.append(t)
        return res

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> res = 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);
                }
            }
            res.add(t);
        }
        return res;
    }
}

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) {
        vector<int> res;
        if (!root)
        {
            return res;
        }

        deque<TreeNode *> deq;
        deq.push_back(root);
        while (!deq.empty())
        {
            int size = deq.size();
            int maxnum = INT_MIN;
            for (int i = 0; i < size; i++)
            {
                TreeNode *ptr = deq.front();
                deq.pop_front();
                if (maxnum < ptr->val)
                {
                    maxnum = ptr->val;
                }

                if (ptr->left)
                {
                    deq.push_back(ptr->left);
                }

                if (ptr->right)
                {
                    deq.push_back(ptr->right);
                }
            }

            res.push_back(maxnum);
        }

        return res;
    }
};

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