105. Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2021-05-26 18:38:00 发布

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本文介绍了一种通过给定的先序遍历和中序遍历序列来构建二叉树的方法。该方法首先找到树根节点，然后递归地构建左子树和右子树。文章提供了一个C++实现示例。

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

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

同第106题，只是树根是先序序列的第一个元素

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    typedef vector<int>::iterator Iter;
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        return buildTreeRecur(inorder.begin(), inorder.end(), preorder.begin(), preorder.end());
    }
    TreeNode *buildTreeRecur(Iter istart, Iter iend, Iter pstart, Iter pend)
    {
        if(istart == iend)return NULL;
        int rootval = *pstart;
        Iter iterroot = find(istart, iend, rootval);
        TreeNode *res = new TreeNode(rootval);
        res->left = buildTreeRecur(istart, iterroot, pstart+1, pstart+1+(iterroot-istart));
        res->right = buildTreeRecur(iterroot+1, iend, pstart+1+(iterroot-istart), pend);
        return res;
    }
};