Leetcode687. 最长同值路径

　　这个路径可能存在从子节点经过父节点，再到子节点的情况，所有从当前节点构成的路径需要考虑左右两条路径相加，用递归，求得左右的最长路径，相加，即为所求

// Study.cpp: 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <string>
#include <algorithm>
#include <sstream>
#include <set>
#include <stack>
#include <iomanip>
#define INT_MAX 2147483647 // maximum (signed) int value
#define INT_MIN (-2147483647 - 1) // minimum (signed) int value
;

using namespace std;

int Max(int a, int b)
{
	return a > b ? a : b;
}

int Min(int a, int b)
{
	return a < b ? a : b;
}
struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}

};
int rmax = 0;
int find_depth(TreeNode* p, int value)
{
 	if (p == nullptr || p->val != value)
		return 0;

	int left = 0, right = 0;
	left = find_depth(p->left,value);
	right = find_depth(p->right,value);

	return 1 + Max(left, right);

}
void solution(TreeNode* p)
{
	if (p == nullptr)
		return;

	int current = find_depth(p->left, p->val) +  find_depth(p->right, p->val);
	if (current > rmax)
		rmax = current;
	solution(p->left);
	solution(p->right);
}

int longestUnivaluePath(TreeNode* root) {
	solution(root);
	return rmax;
}

void printTree(TreeNode* p,int depth=0)
{
	if (p == nullptr)
		return;

	for (int i = 0; i < depth; i++)
		cout << " ";
	cout << p->val << endl;

	printTree(p->left,depth+1);
	printTree(p->right, depth+1);
}
int main()
{
	TreeNode* root = new TreeNode(1);
	root->left = new TreeNode(2);
	root->right = new TreeNode(2);

	TreeNode* p = root->left;
	p->left = new TreeNode(2);
	p->right = new TreeNode(2);
	p->left->left = new TreeNode(2);
	//printTree(p);
	p = root->right;
	p->left = new TreeNode(2);
	p->right = new TreeNode(2);

	printTree(root);
	cout << endl;

	cout << longestUnivaluePath(root);
	system("pause");
	return 0;
}

原文地址：https://www.cnblogs.com/Oscar67/p/9435037.html