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.

这道题跟之前的从中序和后序序列中构造二叉树类似的，换成了从先序和中序序列来构造。

/**
* 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) {
return buildTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
}
private:
TreeNode *buildTree(vector<int> &preorder, int pb, int pe, vector<int> &inorder, int ib, int ie) {
if (pb > pe)
return NULL;

if (pb == pe)
return new TreeNode(preorder[pb]);

int i = ib;
while (i <= ie) {      // 从中序序列中找到前序序列第一个数字（根）的下标
if (inorder[i] == preorder[pb])
break;
++i;
}

TreeNode *root = new TreeNode(preorder[pb]);

if (i > ib)  // 递归构造左子树
root->left = buildTree(preorder, pb+1, pb+i-ib, inorder, ib, i-1);

if (i < ie)   // 递归构造右子树
root->right = buildTree(preorder, pb+i-ib+1, pe, inorder, i+1, ie);

return root;
}
};