给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

先序无法确定子树大小，中序找不到根；所以用先序找根，用中序找大小

题解1 递归

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {unordered_map<int, int> idx;
public:
// 先序自上而下，中序确定左右子树大小TreeNode* build(vector<int>& preorder, vector<int>& inorder, int pre_left, int pre_right, int in_left, int in_right){if(pre_left > pre_right) return nullptr;// 用前序找根（把树单元化，叶子结点看作无左右子的根）， 中序找左树大小int root_idx = pre_left;// 哈希表查此根结点在中序遍历数组的位置int root_in = idx[preorder[root_idx]];TreeNode* root = new TreeNode(preorder[root_idx]);// 左树大小int num_left = root_in - in_left;// left树，越往下走右边界越收紧// 但在preoreder里涉及到赋值，需要往后找，左边界+1root->left = build(preorder, inorder, pre_left+1, pre_left+num_left, in_left, root_in-1);// right树，越往下走左边界约收紧// 同样在preorder里涉及到遍历问题（先序遍历：遍历完左再遍历右，所以到右侧应该是+num_left+1）// 再inorder里就是在根节点的右侧，root_in+1即可root->right = build(preorder, inorder, pre_left+num_left+1, pre_right, root_in+1, in_right);return root;}TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {int s = preorder.size();for(int i = 0; i < s; i++){idx[inorder[i]] = i;}// 含右边界的版本return build(preorder, inorder, 0, s-1, 0, s-1);}
};

题解2 迭代

class Solution {unordered_map<int, int> idx;
public:TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {int s = preorder.size();TreeNode* root = new TreeNode(preorder[0]);stack<TreeNode*> rstk;rstk.push(root);int idx = 0;for(int i = 1; i < s; i++){int val = preorder[i];TreeNode* node = rstk.top();// 判断当前node是不是叶子结点(拐点)if(node->val != inorder[idx]){node->left = new TreeNode(val);rstk.push(node->left);}else{// 用中序查当前先序的结点i是不是右树while(rstk.size() && rstk.top()->val == inorder[idx]){node = rstk.top();rstk.pop();idx ++;}node->right = new TreeNode(val);rstk.push(node->right);}}return root;}
};