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