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].
本题和我之前做过的一个类似的题:Binary Tree Preorder Traversal
https://blog.csdn.net/cherrydreamsover/article/details/81675440
本题是需要我们返回二叉树的后续遍历结果。(左子树–>右子树–>根节点)
代码实现:
/**
* 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> res;
if(root==NULL)
return res;
TreeNode* pNode=root;
TreeNode* prev=root;
stack<TreeNode*> s;
s.push(root);
while(!s.empty())
{
pNode=s.top();
if(prev!=pNode->left && prev!=pNode->right)
{
if(pNode->right)
{
s.push(pNode->right);
}
if(pNode->left)
{
s.push(pNode->left);
}
}
if((pNode->left==NULL && pNode->right==NULL) || prev==pNode->left || prev==pNode->right )
{
res.push_back(pNode->val);
s.pop();
}
prev=pNode;
}
return res;
}
};