勵志用盡量少的代碼做高效的表達。

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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

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題目大意

給定二叉樹的中序和后序遍歷序列，求該二叉樹中根到葉子的路徑和最小者，若存在多個解，則選擇葉子權值最小者（葉子權值不重復）。

思路分析

此題本質是中后序建樹+求路徑最小權。
（PS：書上的數組建樹用不習慣，于是自己寫了一個指針建樹。）
關鍵在于中序和后序建樹，后序遍歷序列的最后一個元素來確定根（前序的話是第一個元素確定根），中序序列來劃分左右子樹，如此遞歸建立樹
對于路徑和求解，建樹完成后對樹進行dfs

#include<bits/stdc++.h> using namespace std; struct Node {int v; //值Node *l = NULL, *r = NULL; //左右子樹 }; vector<int> in, post; //中序，后序存儲 int minsum = 0x3fffff, ans = -1; //路徑最小和、答案(葉子節點) Node* createTree(int i1, int j1, int i2, int j2) {if(i1 >= j1 || i2 >= j2) return NULL; //空樹Node* root = new Node;root->v = post[j2-1]; //根節點為后續遍歷最后一個數int j = find(in.begin()+i1, in.begin()+j1, post[j2-1]) - in.begin();//在中序遍歷中查找該值root->l = createTree(i1, j, i2, i2+(j-i1)); //建立左子樹root->r = createTree(j+1, j1, i2+(j-i1), j2-1); //建立右子樹return root; } void dfs(Node* root, int sum) { //計算到每個葉子的路徑和并記錄最小者if(root->l == NULL && root->r == NULL) { //判斷是否遍歷到頭 sum += root->v;if(sum < minsum || (sum == minsum && ans > root->v)) {minsum = sum;ans = root->v; } return;}if(root->l != NULL) dfs(root->l, root->v+sum); //非空，則訪問左子樹if(root->r != NULL) dfs(root->r, root->v+sum); //非空，則訪問右子樹 } int main() {string s1, s2;int s;while(getline(cin, s1) && getline(cin, s2)) {in.clear(); post.clear(); //初始化stringstream input1(s1), input2(s2);while(input1 >> s) in.push_back(s);while(input2 >> s) post.push_back(s);Node* btree = createTree(0, in.size(), 0, post.size()); //建樹minsum = 0x3fffff; dfs(btree, 0); //遍歷計算printf("%d\n", ans); }return 0; }

收獲：

1、鞏固了DFS
2、掌握了中后序遍歷建樹。
3、掌握了求最小權路徑的方法。

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擇苦而安，擇做而樂，虛擬現實終究比不過真實精彩之萬一。

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