145. 二叉树的后序遍历

难度中等515

给定一个二叉树，返回它的 后序 遍历。

示例:

输入: [1,null,2,3]  1\2/3 输出: [3,2,1]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

1、迭代

/*** 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:vector<int> postorderTraversal(TreeNode* root) {vector<int> res;if(root == nullptr){return res;}stack<TreeNode*> stk;TreeNode* prev = nullptr;while(root != nullptr || !stk.empty()){while(root != nullptr){stk.push(root);root = root->left;}root = stk.top();stk.pop();if(root->right == nullptr || root->right == prev){  //没有右节点res.push_back(root->val);prev = root;root = nullptr;} else{stk.push(root);root = root->right;}}     return res;}
};

2、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 {
public:void backTraverse(TreeNode* root, vector<int>& res){if(root == nullptr){return ;}backTraverse(root->left, res);backTraverse(root->right, res);res.push_back(root->val);}vector<int> postorderTraversal(TreeNode* root) {vector<int> res;backTraverse(root, res);return res;}
};