LeetCode题解： Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

思路：

和inorder+postorder的思路类似。

题解：

/**
 * 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(const vector<int>& preorder, int ps, int pe,
                        const vector<int>& inorder, int is, int ie)
    {
        if (is > ie)
            return nullptr;
            
        int base = preorder[ps];
        TreeNode* node = new TreeNode(base);
        
        int ipos = is;
        while(inorder[ipos] != base) ++ipos;
        
        int left_nodes = ipos - is;
        int right_nodes = ie - ipos;
        
        if (left_nodes != 0)
            node->left = buildTree(preorder, ps + 1, ps + 1 + left_nodes - 1,
                                    inorder, is, is + left_nodes - 1);
        if (right_nodes != 0)
            node->right = buildTree(preorder, ps + 1 + left_nodes, pe,
                                    inorder, ipos + 1, ie);
        
        return node;
    }

    TreeNode *buildTree(const vector<int> &preorder, 
                        const vector<int> &inorder) {
        if (preorder.empty())
            return nullptr;
            
        return buildTree(preorder, 0, preorder.size() - 1,
                        inorder, 0, inorder.size() - 1);
    }
};