'lintcode 具有最大平均数的子树'

Posted on | In

描述

给一棵二叉树，找到有最大平均值的子树。返回子树的根结点

样例

样例 1

输入：
1
/ \
-5 11
/ \ / \
1 2 4 -2
输出：11(它是一个TreeNode)
样例 2

输入：
1
/ \
-5 11
输出：11(它是一个TreeNode)

思考

使用一个函数来遍历这个二叉树。
根据后根遍历的顺序遍历每个节点，用左右子树的和来计算根的和，并且一直保存一个最大的平均数。如果左右子树没有节点，那么根直接返回自己的值，和大小为1。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: the root of binary tree
     * @return: the root of the maximum average of subtree
     */
    TreeNode * findSubtree2(TreeNode * root) {
        // write your code here
        if (!root || (!root->left && !root->right))
            return root;
        double maxAvg = INT_MIN;
        TreeNode *maxNode = NULL;
        int size = 0;
        recur(root,maxNode, size, maxAvg);
        return maxNode;
    }

    double recur(TreeNode *node, TreeNode * &maxNode, int &size, double& maxAvg) {
        if (!node) return 0;
        int leftSize = 0, rightSize = 0;
        double leftAvg,rightAvg;
        leftAvg = recur(node->left, maxNode, leftSize, maxAvg);
        rightAvg = recur(node->right, maxNode, rightSize, maxAvg);
        size = leftSize+rightSize+1;
        double Avg = (leftAvg * leftSize + rightAvg * rightSize + node->val)/(double)size;
        if (Avg > maxAvg) {
            maxAvg = Avg;
            maxNode = node;
        }
        return Avg;

    }
};
-------------end of filethanks for reading-------------