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  • [LeetCode] 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.

    /**
     * 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) {
            int len = inorder.size();
            return build(preorder,0,len-1,inorder,0,len-1);
        }
    private:
        TreeNode * build(vector<int> &preorder,int begpreorder,int endpreorder,vector<int> &inorder,int begInorder,int endInorder){
            
            if(begInorder>endInorder || begpreorder>endpreorder)
                return NULL;
            
            int val = preorder[begpreorder];
            TreeNode *root = new TreeNode(val);
            if(endInorder==begInorder)
                return root;
            int i,j=0;
            for(i=begInorder;i<=endInorder;i++,j++){
               if(inorder[i]==val)
                   break;
            }
                
            root->left = build(preorder,begpreorder+1,begpreorder+j,inorder,begInorder,i-1);
            root->right = build(preorder,begpreorder+1+j,endpreorder,inorder,i+1,endInorder);
            return root;
        
        }
    };
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  • 原文地址:https://www.cnblogs.com/Xylophone/p/3891003.html
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