Construct Binary Tree from Preorder and Inorder Traversal

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/

public TreeNode buildTree(int[] preorder, int[] inorder) {
    Map inMap = new HashMap();

    for(int i = 0; i < inorder.length; i++) {
        inMap.put(inorder[i], i);
    }

    TreeNode root = buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, inMap);
    return root;
}

public TreeNode buildTree(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd, Map inMap) {
    if(preStart > preEnd || inStart > inEnd) return null;

    TreeNode root = new TreeNode(preorder[preStart]);
    int inRoot = inMap.get(root.val);
    int numsLeft = inRoot - inStart;//root的左子节点的数量

    root.left = buildTree(preorder, preStart + 1, preStart + numsLeft, inorder, inStart, inRoot - 1, inMap);
//在前序遍历对应的vector中，如果root有左子节点的话那么对应的为vector中它对应的下一个节点
    root.right = buildTree(preorder, preStart + numsLeft + 1, preEnd, inorder, inRoot + 1, inEnd, inMap);
//在前序遍历对应的vector中，如果root有右节点的话，那么对应的右节点在vector中的索引为preStart + numsLeft + 1(numsLeft 是左节点的数量)
    return root;
}

另外一种不使用递归的算法

class Solution {                                                        
public:
    TreeNode* buildTree(vector &pre, vector &in) {           
           int i = 0, j = 0;
           TreeNode *root = new TreeNode(0x80000000);
           TreeNode *pp = NULL,*ptr = root;
           stack sn;
           sn.push(root);
        
           while (jval == in[j]){//当该节点与 中序遍历的j节点相同说明 当前前序遍历数组中的i(即当前栈顶节点对应的下一个节点)所对应的节点为sn栈顶节点的右节点
                                       //即相当于当前的栈顶节点已经”链接“上了左节点那么下一个索引i对应的节点应为该节点的右节点                      
               pp = sn.top();
                sn.pop();
                j++;
            }else if(pp){
                ptr = new TreeNode(pre[i]);
                pp->right = ptr;
                pp = NULL;
                sn.push(ptr);
                i++;
            }else{//一直左左左
                ptr = new TreeNode(pre[i]);
                sn.top()->left = ptr;
                sn.push(ptr);
                i++;
            }
          }      
          ptr = root->left;
          delete root;
          return ptr;
    }
};