05
04/2015
[LeetCode] Binary Tree Postorder Traversal
Binary Tree Postorder Traversal
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?
解题思路：
1、用递归做很简单，先遍历左孩子，再遍历右孩子，然后遍历父节点。代码如下：
/**
* 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;
postOrder(root, result);
return result;
}
void postOrder(TreeNode *node, vector<int>& result){
if(node==NULL){
return;
}
postOrder(node->left, result);
postOrder(node->right, result);
result.push_back(node->val);
}
};2、但是题目要求不能用递归。此前学长跟我说，若递归改成非递归，无非就是用栈或者队列存储中间结果。这里我们用栈来存储中间结果。注意到，先遍历的后进栈，因此，最先进栈的是父节点，然后是右孩子，最后是左孩子。这里我们需要两个栈，一个存储左孩子栈，一个存储最终结果栈。代码如下：
/**
* 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;
stack<TreeNode*> s; //最终结果栈
stack<TreeNode*> sLeftNode; //左孩子栈
TreeNode* node=root;
while(node!=NULL||!sLeftNode.empty()){
if(node!=NULL){
s.push(node);
if(node->left!=NULL){
sLeftNode.push(node->left);
}
node=node->right;
}else{
node=sLeftNode.top();
sLeftNode.pop();
}
}
while(!s.empty()){
node=s.top();
result.push_back(node->val);
s.pop();
}
return result;
}
};
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