LeetCode Binary Tree PostorderTranversal
Problem Description
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3}
,1
\
2
/
3return
[3,2,1]
.
Note: Recursive solution is trivial, could you do it iteratively?
Problem Solution
1. 递归方案
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
vector<int> nodeVec;
public:
void traverse(TreeNode *root){
if(root==NULL)
return;
traverse(root->left);
traverse(root->right);
nodeVec.push_back(root->val);
}
vector<int> postorderTraversal(TreeNode *root) {
traverse(root);
return nodeVec;
}
};
2. 非递归方案
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
vector<int> nodeVec;
public:void iterTraverse(TreeNode *root){
if(root==NULL)
return;
stack<TreeNode*> st; TreeNode *pCur,*pPrev=NULL; //pCur: current tree node, pPrev: previous visited tree node
st.push(root);
while(!st.empty())
{
pCur=st.top();
if((pCur->left == NULL && pCur->right == NULL) || (pPrev != NULL && (pCur->left==pPrev || pCur->right==pPrev)))
{
nodeVec.push_back(pCur->val);
pPrev=pCur;
st.pop();
}
else
{
if(pCur->right != NULL)
st.push(pCur->right);
if(pCur->left != NULL)
st.push(pCur->left);
}
}
}
vector<int> postorderTraversal(TreeNode *root) {
iterTraverse(root);
return nodeVec;
}
};
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