【LeetCode】105. 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.

有了先序和中序，基本思路就是递归的找左右子树的根节点，加入到树中/** * 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 *root = NULL;  // 存储树的根节点    // 递归函数，pre是先序，pl和pr为左或右子树的区间，in为中序，il和ir同为左或右子树的区间，sub决定是左子树还是右子树，par为父亲结点    void fun(vector<int> &pre, int pl, int pr, vector<int> &in, int il, int ir, int sub, TreeNode *par) {        if (pl <= pr) { // 左或右子树不为空            TreeNode *tmp = new TreeNode(pre[pl]); // 新建一个结点，即某个子树的根节点            if (root == NULL) {                root = tmp; // 根节点                par = root;            } else {                // 根据sub判断添加到左子树还是右子树                if (sub == 0) par->left = tmp;                if (sub == 1) par->right = tmp;            }            // 找到根节点在中序遍历中的位置            int ipos = il;            for (int i = il; i <= ir; ++i) {                if (in[i] == pre[pl]) {                    ipos = i;                    break;                }            }            // 找到先序遍历中左右子树的分隔位置            int ppos = pl + ipos - il;            // 递归构造左右子树            if (ipos > il) fun(pre, pl+1, ppos, in, il, ipos-1, 0, tmp); // 有左子树            if (ipos < ir) fun(pre, ppos+1, pr, in, ipos+1, ir, 1, tmp); // 有右子树        }    }        TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        if (preorder.size() == 0) return NULL;        //TreeNode *root = NULL;        fun(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1, 0, NULL);        return root;    }};