LeetCode Binary Tree Postorder Traversal(二叉树的后序遍历 非递归实现)

题目要求：

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]

.

用栈模拟递归过程， 后序要比其它两个遍历方式稍微复杂些， 要判断左右子节点都遍历后在访问根节点。

代码：

class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
stack<TreeNode*> st;
vector<int> ret;
if(root == NULL)
return ret;
TreeNode* cur = NULL;
TreeNode* pre = NULL;
st.push(root);
while(!st.empty())
{
cur = st.top();
if((cur->left == NULL && cur->right == NULL) ||
(pre != NULL && (cur->left == pre || cur->right == pre)))//如果是叶子节点或者是左右子孩子已经被访问过了，才能访问当前节点
{
ret.push_back(cur->val);
st.pop();
pre = cur;
}
else
{
if(cur->right != NULL)
st.push(cur->right);
if(cur->left != NULL)
st.push(cur->left);
}
}
return ret;

}
};