515. Find Largest Value in Each Tree Row（python+cpp）_python find the largest value in each tree row-优快云博客

本文介绍了一种解决二叉树问题的方法，通过层序遍历找到每层的最大值。提供了Python和C++两种语言的实现代码，适用于算法初学者和面试准备者。

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题目：

You need to find the largest value in each row of a binary tree.
Example:
Input: 

          1
         / \
        3   2
       / \   \  
      5   3   9 

Output: [1, 3, 9]

解释：
返回二叉树每一层的最大值，不就是经典的层序遍历问题？
python代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def largestValues(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        result=[]
        if not root:
            return result
        queue=[root]
        while queue:
            layer_number=len(queue)
            temp_max=float('-inf')
            for i in xrange(layer_number):
                tmp=queue.pop(0)
                if tmp.left:
                    queue.append(tmp.left)
                if tmp.right:
                    queue.append(tmp.right)
                temp_max=max(temp_max,tmp.val)
            result.append(temp_max)
        return result

c++代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include<queue>
using namespace std;
class Solution {
public:
    vector<int> largestValues(TreeNode* root) {
        vector<int>result;
        if(!root)
            return result;
        queue<TreeNode*>_queue;
        _queue.push(root);
        while (!_queue.empty())
        {
            int len_layer=_queue.size();
            int max_val=INT_MIN;
            for (int i=0;i<len_layer;i++)
            {    
                TreeNode* tmp=_queue.front();
                _queue.pop();
                if(tmp->left)
                    _queue.push(tmp->left);
                if(tmp->right)
                    _queue.push(tmp->right);
                max_val=max(max_val,tmp->val); 
            }
            result.push_back(max_val);
        }
        return result; 
    }
};