Binary Tree Postorder Traversal - LeetCode 145

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?

Tags Tree Stack

分析：

二叉树的后序遍历:左右根。

方法一：递归法.

方法二：非递归法，借助栈，新定义一个结构体，标记右子树是否已经访问

以下是C++实现代码，比较简单就不多说了。

/*///递归////0ms////*/
/**
* 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:
void postOrderHelp(TreeNode* root,vector<int> &vec)
{
if(root == NULL)
return;
postOrderHelp(root->left,vec);
postOrderHelp(root->right,vec);
vec.push_back(root->val);

}
vector<int> postorderTraversal(TreeNode* root) {
vector<int> vec;
postOrderHelp(root,vec);
return vec;
}
};

/*//非递归///0ms///*/
/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
struct TreeNodeMod{
TreeNode *node;
bool rightVisited;
};
class Solution {
public:
TreeNodeMod* goFarLeft(TreeNode* root,stack<TreeNodeMod* > &s)
{
if(root == NULL)
return NULL;

TreeNode *cur = root;
TreeNodeMod *newptr;
while(cur ->left != NULL)
{
newptr = new TreeNodeMod();
newptr->node = cur;
newptr->rightVisited = false;
s.push(newptr);
cur = cur->left;
}

newptr = new TreeNodeMod();
newptr->node = cur;
newptr->rightVisited = false;
return newptr;
}
vector<int> postorderTraversal(TreeNode* root) {
stack<TreeNodeMod* > s;
TreeNodeMod *cur = goFarLeft(root,s);
vector<int> vec;
while(cur != NULL)
{
if(cur->node->right == NULL || cur->rightVisited)
{
vec.push_back(cur->node->val);
delete cur;
if(!s.empty())
{
cur = s.top();
s.pop();
}
else
cur = NULL;

}
else
{
cur ->rightVisited = true;
s.push(cur);
cur = goFarLeft(cur->node->right,s);
}
}
return vec;
}
};