https://leetcode-cn.com/problems/binary-tree-postorder-traversal/
思路:非遞歸版,這個思路對於前序遍歷、中序遍歷也適用。左右根,首先獲得棧頂,然後判斷棧頂是否爲空,若不爲空,則再次壓入該節點,同時壓入一個空指針,標記其前一位爲根節點,然後壓入它右節點、左節點;若棧頂爲空,說明前一位也就是當前的棧頂爲根節點,那麼把它的值放進數組中即可。
/**
* 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> ans;
if(root==NULL)
return ans;
stack<TreeNode*> stk;
stk.push(root);
TreeNode *cur;
while(!stk.empty()){
cur=stk.top();
stk.pop();
if(cur){
stk.push(cur);
stk.push(NULL);
if(cur->right)
stk.push(cur->right);
if(cur->left)
stk.push(cur->left);
}
else{
ans.push_back(stk.top()->val);
stk.pop();
}
}
return ans;
}
};