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.

class Solution {
public:
    void f(vector<int> &preorder,int s1, int e1,vector<int> & inorder,int s2, int e2, TreeNode *& root){
        if (s1 > e1){
            return;
        }
        if (!root){
            root = new TreeNode(preorder[s1]);
            int m = s2;
            for(;m <= e2;m++){
                if (inorder[m] == preorder[s1]){
                    break;
                }
            }
            f(preorder,s1 + 1,s1 + 1 + m - 1 - s2,inorder,s2,m-1,root->left);
            f(preorder,e1 - (e2 - m - 1),e1,inorder,m+1,e2,root->right);
        }
    }
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (!preorder.size()){
            return NULL;
        }
        TreeNode * root = NULL;
        f(preorder,0,preorder.size() -1,inorder,0,inorder.size() -1,root);
        return root;        
    }
};

原文链接: https://www.cnblogs.com/kwill/p/3166108.html

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