返回二叉树两个节点间的最大距离

1、题目

给定一棵二叉树的根节点root，任何两个节点之间都存在距离，返回整棵二叉树的最大距离。

2、分析

两个节点之间的距离就是从 A 节点到 B 节点沿途的节点数。

可以有多种解法，但是二叉树套路：

①目标：以 X 为根节点的情况下，整棵树的最大距离；

②可能性：可能和 X 有关，也可能和 X 无关。

• 和X无关的情况有两种可能性：（1）X左树最大距离；（2）X右树最大距离；

• 和X有关的情况的可能性："X左树与X最远 + X右树与X最远 + 1“，而 X 左树与 X 最远就是左树的高度，X右树与X最远就是右树高度，1是X节点。

所以，向左树和右树想获得的信息：（1）树的最大距离；（2）树的高度；

最终结果就是在三种可能性中求最大值。

3、实现

C++ 版

/************************************************************************* > File Name: 037.返回二叉树的最大距离.cpp > Author: Maureen > Mail: [email protected] > Created Time: 五 7/ 1 17:13:32 2022 ************************************************************************/

#include <iostream>
#include <ctime>
#include <unordered_map>
#include <unordered_set>
#include <vector>
using namespace std;

class TreeNode {
    
public:
    int value;
    TreeNode *left;
    TreeNode *right;

    TreeNode(int data) : value(data) {
    }
};

//方法一：暴力枚举，先序遍历获得序列，暴力枚举两个节点之间的距离 
void fillPrelist(TreeNode *root, vector<TreeNode*> &arr) {
    
    if (root == nullptr) return ;

    arr.push_back(root);
    fillPrelist(root->left, arr);
    fillPrelist(root->right, arr);
}

vector<TreeNode *> getPrelist(TreeNode *root) {
    
    vector<TreeNode*> arr;
    fillPrelist(root, arr);
    return arr;
}
//记录每个节点的父节点
void fillParentMap(TreeNode *root, unordered_map<TreeNode*, TreeNode*> &parentMap) {
    
    if (root->left != nullptr) {
    
        parentMap[root->left] = root;
        fillParentMap(root->left, parentMap);
    }

    if (root->right != nullptr) {
    
        parentMap[root->right] = root;
        fillParentMap(root->right, parentMap);
    }
}

unordered_map<TreeNode *, TreeNode *> getParentMap(TreeNode *root) {
    
    unordered_map<TreeNode*, TreeNode*> _map;
    _map[root] = nullptr;
    fillParentMap(root, _map);
    return _map;
}


//获取两个节点之间的距离
int distance(unordered_map<TreeNode*, TreeNode*> &parentMap, TreeNode *o1, TreeNode *o2) {
    
    unordered_set<TreeNode*> o1set;
    TreeNode *cur = o1;
    o1set.insert(cur);

    while (parentMap[cur] != nullptr) {
    
        cur = parentMap[cur];
        o1set.insert(cur);
    }

	//找到o1和o2的最低公共祖先
    cur = o2;
    while (!o1set.count(cur)) {
    
        cur = parentMap[cur];
    }
	
	//统计o1和o2到最低公共祖先的距离
    TreeNode *lowestAncestor = cur;
    cur = o1;
    int distance1 = 1;
    while (cur != lowestAncestor) {
    
        cur = parentMap[cur];
        distance1++;
    }

    cur = o2;
    int distance2 = 1;
    while (cur != lowestAncestor) {
    
        cur = parentMap[cur];
        distance2++;
    }
    
    return distance1 + distance2 - 1;
}

int maxDistance1(TreeNode *root) {
    
    if (root == nullptr) return 0;

    vector<TreeNode*> arr = getPrelist(root);

    unordered_map<TreeNode *, TreeNode *> parentMap = getParentMap(root);

    int ans = 0;
    for (int i = 0; i < arr.size(); i++) {
    
        for (int j = i; j < arr.size(); j++) {
    
            ans = max(ans, distance(parentMap, arr[i], arr[j]));
        }
    }
    return ans;
}

//方法二 
class Info {
    
public:
    int maxDistance;
    int height;

    Info(int m, int h) {
    
        maxDistance = m;
        height = h;
    }
};


Info *process(TreeNode *x) {
    
    if (x == nullptr) {
    
        return new Info(0, 0);
    }

    Info *leftInfo = process(x->left);
    Info *rightInfo = process(x->right);

    int height = max(leftInfo->height, rightInfo->height) + 1;

    int p1 = leftInfo->maxDistance;
    int p2 = rightInfo->maxDistance;
    int p3 = leftInfo->height + rightInfo->height + 1;

    int maxDistance = max(max(p1, p2), p3);

    return new Info(maxDistance, height);
}

int maxDistance2(TreeNode *root) {
    
    return process(root)->maxDistance;
}


//for test
TreeNode *generate(int level, int maxl, int maxv) {
    
    if (level > maxl || (rand() % 100 / (double)101) < 0.5)
        return nullptr;

    TreeNode *root = new TreeNode(rand() % maxv);
    root->left = generate(level + 1, maxl, maxv);
    root->right = generate(level + 1, maxl, maxv);

    return root;
}

TreeNode *generateRandomBST(int level, int value) {
    
    return generate(1, level, value);
}

int main() {
    
    srand(time(0));

    int level = 5;
    int value = 100;
    int testTimes = 1000001;

    cout << "Test begin:" << endl;
    for (int i = 0; i < testTimes; i++) {
    
        TreeNode *root = generateRandomBST(level, value);
        if (maxDistance1(root) != maxDistance2(root)) {
    
            cout << "Failed!" << endl;
            return 0;
        }
        if (i && i % 100 == 0) cout << i << " cases passed!" << endl;
    }
    cout << "Success!" << endl;
    return 0;
}

Java 版

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;

import javax.security.auth.x500.X500PrivateCredential;

public class MaxDistance {
    
    public static class Node {
    
		public int value;
		public Node left;
		public Node right;

		public Node(int data) {
    
			this.value = data;
		}
	}

    //方法一：暴力枚举，先序遍历获得序列，暴力枚举两个节点之间的距离
	public static int maxDistance1(Node head) {
    
		if (head == null) {
    
			return 0;
		}
		ArrayList<Node> arr = getPrelist(head);
		HashMap<Node, Node> parentMap = getParentMap(head);
		int max = 0;
		for (int i = 0; i < arr.size(); i++) {
    
			for (int j = i; j < arr.size(); j++) {
    
				max = Math.max(max, distance(parentMap, arr.get(i), arr.get(j)));
			}
		}
		return max;
	}

	public static ArrayList<Node> getPrelist(Node head)
		ArrayList<Node> arr = new ArrayList<>();
		fillPrelist(head, arr);
		return arr;
	}

	public static void fillPrelist(Node head, ArrayList<Node> arr) {
    
		if (head == null) {
    
			return;
		}
		arr.add(head);
		fillPrelist(head.left, arr);
		fillPrelist(head.right, arr);
	}

	public static HashMap<Node, Node> getParentMap(Node head) {
    
		HashMap<Node, Node> map = new HashMap<>();
		map.put(head, null);
		fillParentMap(head, map);
		return map;
	}

	public static void fillParentMap(Node head, HashMap<Node, Node> parentMap) {
    
		if (head.left != null) {
    
			parentMap.put(head.left, head);
			fillParentMap(head.left, parentMap);
		}
		if (head.right != null) {
    
			parentMap.put(head.right, head);
			fillParentMap(head.right, parentMap);
		}
	}

	public static int distance(HashMap<Node, Node> parentMap, Node o1, Node o2) {
    
		HashSet<Node> o1Set = new HashSet<>();
		Node cur = o1;
		o1Set.add(cur);
		while (parentMap.get(cur) != null) {
    
			cur = parentMap.get(cur);
			o1Set.add(cur);
		}
		cur = o2;
		while (!o1Set.contains(cur)) {
    
			cur = parentMap.get(cur);
		}
		Node lowestAncestor = cur;
		cur = o1;
		int distance1 = 1;
		while (cur != lowestAncestor) {
    
			cur = parentMap.get(cur);
			distance1++;
		}
		cur = o2;
		int distance2 = 1;
		while (cur != lowestAncestor) {
    
			cur = parentMap.get(cur);
			distance2++;
		}
		return distance1 + distance2 - 1;
	}

    //方法二：递归套路
    public static int maxDistance2(Node head) {
    
        return process(head).maxDistance;
    }

    public static class Info {
    
        public int maxDistance; //树的最大距离
        public int height; //树的高度

        public Info(int m, int h) {
    
            this.maxDistance = m;
            this.height = h;
        }
    }

    //一定要返回Info，否则无法约束子树
    public static Info process(Node x) {
    
        //空树距离为0，值是好设置的
        if (x == null) return new Info(0, 0);

        //左树的信息
        Info leftInfo = process(x.left);

        //右树的信息
        Info rightInfo = process(x.right);

        //根据左右树的信息得出当前节点的信息
        //当前节点为根的树的高度：max(左树高度，右树高度) + 1，1是当前节点这一层
        int height = Math.max(leftInfo.height, rightInfo.height) + 1;

        //可能性1：不经过x节点，单纯使用左树的最大距离
        int p1 = leftInfo.maxDistance;
        //可能性2：不经过x节点，单纯使用右树的最大距离
        int p2 = rightInfo.maxDistance;
        //可能性3：要经过x节点，左树的高度 + 右树的高度 + 1（自己）
        int p3 = leftInfo.height + rightInfo.height + 1;

        int maxDistance = Math.max(Math.max(p1, p2), p3);

        return new Info(maxDistance, height);
    }


    //对数器
    // for test
	public static Node generateRandomBST(int maxLevel, int maxValue) {
    
		return generate(1, maxLevel, maxValue);
	}

	// for test
	public static Node generate(int level, int maxLevel, int maxValue) {
    
		if (level > maxLevel || Math.random() < 0.5) {
    
			return null;
		}
		Node head = new Node((int) (Math.random() * maxValue));
		head.left = generate(level + 1, maxLevel, maxValue);
		head.right = generate(level + 1, maxLevel, maxValue);
		return head;
	}

	public static void main(String[] args) {
    
		int maxLevel = 4;
		int maxValue = 100;
		int testTimes = 1000000;
		for (int i = 0; i < testTimes; i++) {
    
			Node head = generateRandomBST(maxLevel, maxValue);
			if (maxDistance1(head) != maxDistance2(head)) {
    
				System.out.println("Oops!");
			}
		}
		System.out.println("finish!");
	}
}