面试题15：有前或者后续遍历和中序遍历构建二叉树

二叉树的构建必须要有中序遍历，另外还需要前序和后续中的一种。下面分别是有前序和中序、后续和中序采用递归方法构建二叉树的过程

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.empty()) return nullptr;
        int n = preorder.size();
        int rootVal = preorder[0];

        TreeNode* root = new TreeNode(rootVal);
        vector<int> leftIn(inorder.begin(),find(inorder.begin(),inorder.end(),rootVal));
        int leftSize = leftIn.size();
        vector<int> leftPre(preorder.begin()+1,preorder.begin()+1+leftSize);

        vector<int> rightIn(inorder.begin()+leftSize+1,inorder.end());
        vector<int> rightPre(preorder.begin()+leftSize+1,preorder.end());
        root->left= buildTree(leftPre,leftIn);
        root->right = buildTree(rightPre,rightIn);

        return root;
    }
};

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution{
public:
    TreeNode* buildTree(vector<int>& inorder,vector<int>& postOrder){
        if(postOrder.empty()) return nullptr;

        int n = postOrder.size();
        int rootVal = postOrder[n-1];
        TreeNode* root = new TreeNode(rootVal);

        vector<int> leftIn(inorder.begin(),find(inorder.begin(),inorder.end(),rootVal));
        int leftSize = leftIn.size();
        vector<int> leftPost(postOrder.begin(),postOrder.begin()+leftSize);

        vector<int> rightIn(inorder.begin()+leftSize+1,inorder.end());
        vector<int> rightPost(postOrder.begin()+leftSize,postOrder.end()-1);

        root->left = buildTree(leftIn,leftPost);
        root->right = buildTree(rightIn,rightPost);
        return root;
    }
};