栈

• 思路：
  • 思路与 力扣105. 从前序与中序遍历序列构造二叉树 相同；
  • 差异的地方：
    • 从后序遍历数组尾部向前遍历；（根节点在尾部）
    • 一直迭代“最右”节点，将其挂载到栈顶（“根”节点）的右子树节点；（后序遍历从尾部迭代顺序变成了：根-右子树-左子树）
    • 出栈后，挂载左子树；

/*** 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 {
public:TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {if (!postorder.size()) {return nullptr;}int post_size = postorder.size();std::stack<TreeNode*> stk;TreeNode* root = new TreeNode(postorder[post_size - 1]);stk.push(root);int in_idx = post_size - 1;for (int idx = post_size - 2; idx >= 0; --idx) {TreeNode *node = stk.top();int post_val = postorder[idx];if (node->val != inorder[in_idx]) {node->right = new TreeNode(post_val);stk.push(node->right);} else {while (!stk.empty() && (stk.top()->val == inorder[in_idx])) {node = stk.top();stk.pop();--in_idx;}node->left = new TreeNode(post_val);stk.push(node->left);}}return root;}
};