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;
}
};