145. Binary Tree Postorder Traversal
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?
非递归实现二叉树的后序遍历
/**
* 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) {
stack<TreeNode*> sta;
TreeNode* cur;
TreeNode* pre=NULL;
vector<int> vec;
if (root==NULL)
{
return vec;
}
if (root->left == NULL&&root->right == NULL)
{
vec.push_back(root->val);
return vec;
}
else
{
sta.push(root); //保证左节点先于右节点被访问，根节点后于左右节点被访问
while (!sta.empty())
{
cur = sta.top();
if ((cur->left==NULL&&cur->right==NULL)||pre!=NULL&&(pre==cur->left||pre==cur->right))
{ //访问该节点
vec.push_back(cur->val);
sta.pop();
pre = cur;
}
else
{
if (cur->right)
{
sta.push(cur->right);
}
if (cur->left)
{
sta.push(cur->left);
}
}
}
}
return vec;
}
};
