题目1509：树中两个结点的最低公共祖先

后续遍历即可，这样每次处理当前节点时，可以综合左右节点的信息进行处理。时间复杂度为O(n)

PS : 经查找维基百科LCA的定义，一个节点可以称为他自己的子节点。

注意：以下代码九度上第五个case没过

#include<iostream>

using namespace std;

struct Node{

	int val;

	Node * left;

	Node * right;

	Node (int value){

			val = value;

			left  = NULL;

			right = NULL;

	}

};

Node * createTree(){

	int value;

	cin>>value;

	if(value == 0) return NULL;

	Node *root = new Node(value);

	root->left = createTree();

	root->right = createTree();

	return root;

}

void destroy(Node * root){

	if(root == NULL) return ;

	Node * l = root->left;

	Node * r = root->right;

	delete root;

	destroy(l);

	destroy(r);

}

int findLCA(Node * root, int m1, int m2, int &value){



		if(root == NULL || m1 == 0 || m2 == 0) return 0;

		int findleft = 0, findright = 0, mid = 0;

		findleft = findLCA(root->left, m1, m2, value);

		if(findleft < 2)

		  findright = findLCA(root->right, m1, m2, value);

		

		if(root->val == m1 && root->val == m2) 

				mid = 2;

			else if(root->val == m1 || root->val == m2) 

					mid = 1;

		

		if(findleft == 1 && findright == 1 ){

			value = root->val;

			return 2;

		}

		if((findleft == 1 || findright == 1) && mid == 1){

			value = root->val;

			return 2;

		}

		if(mid == 2){

		    value = root->val;

			return 2;

		}

		

		return findleft | findright | mid;

}



int main(){



	int n;

	int m1, m2, value;

	

	cin>>n;

	while(n--){

		Node * root = createTree();

	    cin>>m1>>m2;

		value = 0;

		findLCA(root, m1, m2, value);

		if(value != 0) 

		    cout<<value<<endl;

	    else

		 cout<<"My God"<<endl;

		 

		destroy(root);

	}

	return 0;

}

　　ps 贴一个九度内部AC的代码 ，花了我20个九度豆买的，奶奶的

#include<stdio.h>

#define N 10005

int tree[N][4],e;

int find(int x){

    int i;

    for(i=0;i<=e;i++)

        if(tree[i][3]==x)

            return i;

    return -1;

}

void build(int a)

{

    int temp;

    scanf("%d",&temp);

    if(temp)

    {

        tree[a][1]=e;

        tree[e][3]=temp;

        tree[e][0]=a;

        e++;

        build(e-1);

    }

    scanf("%d",&temp);

    if(temp)

    {

        tree[a][2]=e;

        tree[e][3]=temp;

        tree[e][0]=a;

        e++;

        build(e-1);

    }

}

int main()

{

    int n,i,m1,m2,s1,s2,t1,t2,c1,c2,c;

    while(scanf("%d",&n)!=EOF)

    {

        while(n--)

        {

            e=1;

            scanf("%d",&tree[0][3]);

            tree[0][0]=-1;

            build(0);

            scanf("%d %d",&m1,&m2);

            s1=s2=-1;

            s1=find(m1);

            s2=find(m2);

            if(s1!=-1&&s2!=-1)

            {

                t1=s1;

                t2=s2;

                c1=c2=0;

                while(tree[t1][0]!=-1)

                {

                    c1++;

                    t1=tree[t1][0];

                }

                while(tree[t2][0]!=-1)

                {

                    c2++;

                    t2=tree[t2][0];

                }

                if(c1>c2)

                {

                    c=c1-c2;

                    while(c--)

                        s1=tree[s1][0];

                }

                if(c2>c1)

                {

                    c=c2-c1;

                    while(c--)

                        s2=tree[s2][0];

                }

                while(s1!=s2)

                {

                    s1=tree[s1][0];

                    s2=tree[s2][0];

                }

                printf("%d\n",tree[s1][3]);

            }

            else

                printf("My God\n");

        }

    }

    return 0;

}

/**************************************************************

    Problem: 1509

    User: 282544883

    Language: C

    Result: Accepted

    Time:90 ms

    Memory:1068 kb

****************************************************************/