题目描述

根据一棵树的前序遍历与中序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。

样例

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]

返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

算法1

(递归) O(n^2)

C 代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */


struct TreeNode* dfs(int *pre, int *in, int pl, int pr, int il, int ir, int n) 
{
    if (pl > pr) {
        return NULL;
    }
    int k = 0;
    for (int i = 0; i < n; i ++ ) {
        if (in[i] == pre[pl]) {
            k = i - il;
        }
    }
    struct TreeNode *root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    memset(root, 0x00, sizeof(struct TreeNode));
    root->val = pre[pl];
    root->left = dfs(pre, in, pl + 1, pl + k, il, il + k + 1, n);
    root->right = dfs(pre, in, pl + k + 1, pr, il + k + 1, ir, n);
    return root;
}

struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize){
    int n = preorderSize;
    return dfs(preorder, inorder, 0, preorderSize - 1, 0, inorderSize - 1, n);
}