题目描述
Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root.
Note: The length of path between two nodes is represented by the number of edges between them.
方法思路
对于这种树的路径问题,递归是不二之选。在递归函数中,我们首先对其左右子结点调用递归函数,得到其左右子树的最大相同值路径,下面就要来看当前结点和其左右子结点之间的关系了,如果其左子结点存在且和当前节点值相同,则left自增1,否则left重置0;同理,如果其右子结点存在且和当前节点值相同,则right自增1,否则right重置0。然后用left+right来更新结果res。而调用当前节点值的函数只能返回left和right中的较大值,因为如果还要跟父节点组path,就只能在左右子节点中选一条path,当然选值大的那个了
作者:随波逐流he
来源:CSDN
原文:https://blog.csdn.net/qq_36734094/article/details/78507945
版权声明:本文为博主原创文章,转载请附上博文链接!
class Solution {
//Runtime: 4 ms, faster than 100.00%
//Memory Usage: 41 MB, less than 97.34%
int len = 0; // global variable
public int longestUnivaluePath(TreeNode root) {
if (root == null) return 0;
len = 0;
getLen(root, root.val);
return len;
}
private int getLen(TreeNode node, int val) {
if (node == null) return 0;
int left = getLen(node.left, node.val);
int right = getLen(node.right, node.val);
len = Math.max(len, left + right);
if (val == node.val)
return Math.max(left, right) + 1;
return 0;
}
}