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  • Construct Binary Tree from Preorder and Inorder Traversal

    Construct Binary Tree from Preorder and Inorder Traversal

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

    分析: 根据前序遍历和中序遍历构造一棵树,递归求解即可

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
        void Build(int l1,int l2, int r1,int r2, const vector<int>& pre, const vector<int> & in, TreeNode*& root){
            root = new TreeNode(pre[l1]);
            int i;
            for(i=r1; i<=r2; i++)
                if(in[i]==root->val)
                    break;
            if(i==r1)
                root->left=nullptr;
            else
                Build(l1+1, l1+i-r1,r1, i-1 ,pre,in, root->left);
            if(i==r2)
                root->right=nullptr;
            else
                Build(l1+i-r1+1, l2, i+1, r2, pre, in, root->right);
            return;
        }
        TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
            if(preorder.size()==0 && inorder.size()==0)
                return nullptr;
            TreeNode* root;
            Build(0,preorder.size()-1, 0, inorder.size()-1, preorder, inorder, root);
            return root;
        }
    };
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  • 原文地址:https://www.cnblogs.com/willwu/p/6056730.html
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