[LeetCode]108. Construct Binary Tree from Preorder and Inorder Traversal由前序序列和中序序列重建二叉树

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

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

 

解法：前序遍历的第一个元素为树根，因为树中无重复元素，因此遍历一遍可以在中序序列中找出树根所在位置，于是把中序序列分成前后两部分，分别为树根的左右子树。再递归调用重建左右子树即可。注意确定左右子树时中序序列和后序序列的两个下标值。rootIndex只能确定中序序列中左右子树的元素，而不能确定前序序列的左右子树元素位置，后者需要根据左右子树元素个数来确定：第一个为树根，接下来的leftLength个为左子树元素，最后rightLength个为右子树元素。

/**
 * 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:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int pn = preorder.size(), in = inorder.size();
        if (pn == 0 || in == 0 || pn != in) return NULL;
        return buildRecursion(preorder, 0, pn - 1, inorder, 0, in - 1);
    }
private:
    TreeNode* buildRecursion(vector<int>& preorder, int pbeg, int pend, vector<int>& inorder, int ibeg, int iend) {
        TreeNode* root = new TreeNode(preorder[pbeg]);
        if (ibeg == iend) return root;
        int rootIndex = ibeg;
        while (rootIndex <= iend && inorder[rootIndex] != root->val)
            ++rootIndex;
        int leftLength = rootIndex - ibeg, rightLength = iend - rootIndex;
        if (leftLength > 0)
            root->left = buildRecursion(preorder, pbeg + 1, pbeg + leftLength, inorder, ibeg, rootIndex - 1);
        if (rightLength > 0)
            root->right = buildRecursion(preorder, pbeg + leftLength + 1, pend, inorder, rootIndex + 1, iend);
        return root;
    }
};