【题目】

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?

【题意】

非递归实现兴许遍历

【思路】

维护两个栈,一个栈用来存储标记,标记对应的结点的右子树是否已经被遍历。还有一个栈存储树节点,用以模拟后序遍历。

【代码】

/**
* 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>result;
TreeNode*node=root;
stack<TreeNode*>st;
stack<bool>st_flag; while(node){
st.push(node);
st_flag.push(false);
node=node->left;
}
while(!st.empty()){
bool status = st_flag.top();
if(!status){
//訪问右子树
st_flag.pop(); st_flag.push(true);
node = st.top();
node=node->right;
while(node){
st.push(node);
st_flag.push(false);
node=node->left;
}
}
else{
//訪问当前结点
st_flag.pop();
node = st.top(); st.pop();
result.push_back(node->val);
}
}
return result;
}
};

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