UVa 548 Tree(建樹,遞歸遍歷)

題意 給你一個樹的中序遍歷和後序遍歷 某個節點的權值為從根節點到該節點所經過節點的和 求權值最小的葉節點的值 如果存在多個 輸出值最小的那個

把樹建好就好說了 遞歸遞歸dfs msun保存最小葉節點權值 ans保存答案

#include
#include
#include
using namespace std;
const int maxn = 10005, INF = 0x3f3f3f3f;
int in[maxn], post[maxn], msum, ans;

struct node
{
    int val;
    node *left, *right;
    node(): left(NULL), right(NULL) {}
};

node* build(int le, int ri, int pos)
{
    node *cur = new node;
    cur->val = post[pos];
    if(le == ri) return cur;
    int i = le;
    while(in[i] != post[pos]) ++i;
    if(i == le) cur->left = NULL;
    else  cur->left = build(le, i - 1, pos - 1 - (ri - i));
    if(i == ri) cur->right = NULL;
    else cur->right = build(i + 1, ri, pos - 1);
    return cur;
}

void dfs(node *cur, int sum)
{
    int t = cur->val;
    if(!(cur->left || cur->right))
    {
        if(sum + t < msum || (sum + t == msum && t < ans))
            msum = sum + t, ans = t;
    }
    if(cur->left != NULL) dfs(cur->left, sum + t);
    if(cur->right != NULL) dfs(cur->right, sum + t);

}

int main()
{
    char s[maxn*10];
    while(gets(s) != NULL)
    {
        int t = 0, l = 0;
        for(int i = 0; s[i] != '\0'; ++i)
            if(isdigit(s[i])) t = t * 10 + s[i] - '0';
            else  in[++l] = t, t = 0;
        in[++l] = t;

        gets(s);
        l = t = 0;
        for(int i = 0; s[i] != '\0'; ++i)
            if(isdigit(s[i])) t = t * 10 + s[i] - '0';
            else post[++l] = t, t = 0;
        post[++l] = t;

        node *root = build(1, l, l);
        ans = msum = INF;
        dfs(root, 0);
        printf("%d\n", ans);
    }
    return 0;
}

Tree

You are to determine the value of the leaf node in a given binary tree that is the terminal node of a path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values of nodes along that path.

Input

The input file will contain a description of the binary tree given as the inorder and postorder traversal sequences of that tree. Your program will read two line (until end of file) from the input file. The first line will contain the sequence of values associated with an inorder traversal of the tree and the second line will contain the sequence of values associated with a postorder traversal of the tree. All values will be different, greater than zero and less than 10000. You may assume that no binary tree will have more than 10000 nodes or less than 1 node.

Output

For each tree description you should output the value of the leaf node of a path of least value. In the case of multiple paths of least value you should pick the one with the least value on the terminal node.

Sample Input

3 2 1 4 5 7 6
3 1 2 5 6 7 4
7 8 11 3 5 16 12 18
8 3 11 7 16 18 12 5
255
255

Sample Output

1
3
255

Tree

You are to determine the value of the leaf node in a given binary tree that is the terminal node of a path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values of nodes along that path.

Input

The input file will contain a description of the binary tree given as the inorder and postorder traversal sequences of that tree. Your program will read two line (until end of file) from the input file. The first line will contain the sequence of values associated with an inorder traversal of the tree and the second line will contain the sequence of values associated with a postorder traversal of the tree. All values will be different, greater than zero and less than 10000. You may assume that no binary tree will have more than 10000 nodes or less than 1 node.

Output

For each tree description you should output the value of the leaf node of a path of least value. In the case of multiple paths of least value you should pick the one with the least value on the terminal node.

Sample Input

3 2 1 4 5 7 6
3 1 2 5 6 7 4
7 8 11 3 5 16 12 18
8 3 11 7 16 18 12 5
255
255

Sample Output

1
3
255