105. Construct Binary Tree from Preorder and Inorder Traversal

题目

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

分析

《剑指Offer》上的重建二叉树原题，根据前序序列是根节点，中序序列当前根节点左侧全是左子树，右侧全是右子树的特点重建，需要注意的是，要计算出左子树长度，并在前序序列中截取出左子树和右子树的部分。

class Solution {
public:
    TreeNode* build(vector& preorder, int pre_beg, int pre_end, vector& inorder, int in_beg, int in_end){
        if(pre_beg>=pre_end||in_beg>=in_end){
            return NULL;
        }
        int i;
        for(i=in_beg;i0){//如果左子树存在，则在前序序列当前位置的后leftLen个属于左子树
            root->left=build(preorder,pre_beg+1,pre_beg+leftLen+1,inorder,in_beg,i);
        }
        if(leftLenright=build(preorder,pre_beg+leftLen+1,pre_end,inorder,i+1,in_end);//前序和中序区间按照[beg,end)截取
        }
        
        return root;
        
    }
    TreeNode* buildTree(vector& preorder, vector& inorder) {
        if(preorder.size()!=inorder.size()||preorder.empty()||inorder.empty())
            return NULL;
        return build(preorder,0,preorder.size(),inorder,0,inorder.size());
    }
};