LeetCode 145 Binary Tree Postorder Traversal（二叉树后序遍历）

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?

题目大意：写出二叉树非递归后序遍历的方法。

解题思路：对于任意节点N：

    1. 将N入栈

    2. 如果N没有左孩子和右孩子，或者N存在左孩子或右孩子但都已经被访问过了，那么可以直接访问N

    3. 如果N存在左孩子或右孩子且还没被访问过，那么将N的右孩子和左孩子依次入栈，然后将栈顶元素置为N，重复步骤2和3

    4. 直到N为空且栈为空，遍历结束

参考资料：http://www.cnblogs.com/dolphin0520/archive/2011/08/25/2153720.html

代码如下:

/**
* 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) {
if(root == nullptr) return vector<int>();
vector<int> ans;
stack<TreeNode*> stk;
TreeNode* cur;
TreeNode* pre = nullptr;
stk.push(root);

while(stk.size()){
cur = stk.top();
if((cur->left == nullptr && cur->right == nullptr) || (pre && (pre == cur->left || pre == cur->right))){
ans.push_back(cur->val);
stk.pop();
pre = cur;
}else{
if(cur->right) stk.push(cur->right);
if(cur->left) stk.push(cur->left);
}
}
return ans;
}
};