题目描述
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},
1 \ 2 / 3
return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
非递归实现:
语言:C++ 运行时间: <1 ms 占用内存:8568K 状态:答案正确
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/** *
Definition for binary tree *
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 >
v; stack<TreeNode*>
s; map<TreeNode*, int >mp; //构造
节点指针,访问flag 的map while
(root != NULL || !s.empty()){ while
(root != NULL){ s.push(root); mp[root]
= 1 ; //怎么记录flag???? root
= root->left; } while
(!s.empty() && mp[s.top()]== 2 ){ v.push_back(s.top()->val); s.pop(); } if
(!s.empty()){ mp[s.top()]
= 2 ; root
= s.top()->right; } } return
v; } }; |
/**
*
Definition for binary tree
*
struct TreeNode {
*
int val;
*
TreeNode *left;
*
TreeNode *right;
*
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
*
};
*/
//=====================递归方法is
trivial。不过还是写出来吧=================================
class
Solution {
public
:
void
postOrder(TreeNode *root,vector<
int
>&vec){
if
(root
!= NULL){
postOrder(root->left,vec);
postOrder(root->right,vec);
vec.push_back(root->val);
}
}
vector<
int
>
postorderTraversal(TreeNode *root) {
vector<
int
>vec;
postOrder(root,vec);
return
vec;
}
};