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题目描述
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?
解题思路1:
用两个栈和一个数组。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
stack<TreeNode*> s1;
stack<TreeNode*> s2;
vector<int> res;
if(!root){
return res;
}
s1.push(root);
TreeNode* p=root;
while(!s1.empty()){
p=s1.top();
s1.pop();
s2.push(p);
if(p->left!=NULL){
s1.push(p->left);
}
if(p->right!=NULL){
s1.push(p->right);
}
}
while(!s2.empty()){
p=s2.top();
res.push_back(p->val);
s2.pop();
}
return res;
}
};
解题思路2:
用一个栈和一个数组。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
stack<TreeNode*> s;
TreeNode* p=root,* last_visit=root;
vector<int> res;
while(p!=NULL||!s.empty()){
while(p!=NULL){
s.push(p);
p=p->left;
}
p=s.top();
if(p->right==NULL||last_visit==p->right){
res.push_back(p->val);
last_visit=p;
s.pop();
p=NULL;
}
else{
p=p->right;
}
}
return res;
}
};
递归:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
vector<int> out;
postorder(root, out);
return out;
}
void postorder(TreeNode* root, vector<int>& out) {
if(root == NULL)
return;
postorder(root->left, out);
postorder(root->right, out);
out.push_back(root -> val);
}
};