【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】
非递归实现后续遍历【思路】
维护两个栈,一个栈用来存储标记,标记相应的结点的右子树是否已经被遍历;另一个栈存储树节点,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){
//访问右子树
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{
//访问当前结点
st_flag.pop();
node = st.top(); st.pop();
result.push_back(node->val);
}
}
return result;
}
};