二叉树前序、中序和后序遍历相互求法

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假设已知一个二叉树，其前序遍历为1,2,4,7,3,5,6,8；中序遍历为4,7,2,1,5,3,8,6;后序遍历为7, 4, 2, 5, 8, 6, 3, 1。
在已知前序遍历和中序遍历或者是中序遍历和后序遍历两者时，其二叉树顺序是确定的，即另一遍序方式是确定的。
在求解过程中，首先需要确定根节点（前序遍历的第一个点为根节点，后序遍历最后一个点为根节点），然后再在中序遍历中寻找根节点，故而确定左右子树，最后利用递归的方式即可得出结果。

#include <iostream>#include <vector>using namespace std;struct TreeNode {   int val;   TreeNode *left;   TreeNode *right;   TreeNode(int x) : val(x), left(NULL), right(NULL) {}}*TNode;//已知前序和中序struct TreeNode* reConstructBinaryTreePre(vector<int> pre, vector<int> in);//已知中序和后序struct TreeNode* reConstructBinaryTreePost(vector<int> in, vector<int> post);void PostOrderTraverse(TreeNode *rt);void PreOrderTraverse(TreeNode *rt);int main(){int ppre[] = {1,2,4,7,3,5,6,8};int iin[] = {4,7,2,1,5,3,8,6};int ppos[] = {7, 4, 2, 5, 8, 6, 3, 1};vector<int> pre;vector<int> in;vector<int> post;for(int i = 0; i < 8; i++){    pre.push_back(ppre[i]);    in.push_back(iin[i]);    post.push_back(ppos[i]);}cout << "后序遍历：" << endl;TNode = reConstructBinaryTreePre(pre, in);PostOrderTraverse(TNode);cout << endl;cout << "前序遍历：" << endl;TNode = reConstructBinaryTreePost(in, post);PreOrderTraverse(TNode);cout << endl;return 0;}//后序遍历void PostOrderTraverse(TreeNode *rt){if(rt != NULL){    PostOrderTraverse(rt -> left);    PostOrderTraverse(rt -> right);    cout << rt -> val;}}//前序遍历void PreOrderTraverse(TreeNode *rt){if(rt != NULL){    cout << rt -> val;    PreOrderTraverse(rt -> left);    PreOrderTraverse(rt -> right);}}//已知前序遍历和中序遍历，重建二叉树struct TreeNode* reConstructBinaryTreePre(vector<int> pre,vector<int> in) {    TreeNode* post = (TreeNode*)malloc(sizeof(struct TreeNode));    int len = pre.size();    if(len <= 0)        return NULL;    post -> val = pre[0];    vector<int>::iterator iter = find(in.begin(), in.end(), pre[0]);    vector<int> lpre(pre.begin() + 1, pre.begin() + (iter - in.begin()) + 1);    vector<int> lin(in.begin(), iter);    post -> left = reConstructBinaryTreePre(lpre, lin);    vector<int> rpre(pre.begin() + (iter - in.begin()) + 1, pre.end());    vector<int> rin(iter + 1, in.end());    post -> right = reConstructBinaryTreePre(rpre, rin);    return post;}//已知中序遍历和后序遍历，构建二叉树struct TreeNode* reConstructBinaryTreePost(vector<int> in, vector<int> post){TreeNode* pre = (TreeNode*)malloc(sizeof(struct TreeNode));int len = in.size();if(len <= 0)    return NULL;pre -> val = post[len - 1];vector<int>::iterator iter = find(in.begin(), in.end(), post[len - 1]);vector<int> lin(in.begin(), iter);vector<int> lpost(post.begin(), post.begin() + (iter - in.begin()) + 1);pre -> left = reConstructBinaryTreePost(lin, lpost);vector<int> rin(iter + 1, in.end());vector<int> rpost(post.begin() + (iter - in.begin()), post.end() - 1);pre -> right = reConstructBinaryTreePost(rin, rpost);return pre;}

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