[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;    }};