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?

思路一：直接使用递归方法

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void postorder(TreeNode *root,vector<int> &data)
    {
        if(root==NULL)
            return;
        postorder(root->left,data);
        postorder(root->right,data);
        data.push_back(root->val);
    }
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> result;
        postorder(root,result);
        return result;
    }
};

思路二：循环方法。后序遍历使用栈来做暂时的存储，而且树的后序遍历相对于树的前序遍历和中序遍历比较复杂，处理起来比较麻烦，一般这样我们都会使用一个pre变量来保存上次输出的结点，cur为当前结点，如果pre正好是cur的左子树或者右子树，那么可以输出cur，否则，说明还有cur的左子树或者右子树等待输出，也就是还没到cur输出的时候。另外cur没有左右子树即叶结点，可以输出。但是下面代码使用反向思维，先依次输出根结点->右子树结点->左子树结点，存到result，然后反转就是后序遍历的结果。

 

/**
 * 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;
        if(root==NULL)
            return result;
        stack<TreeNode*> sTree;
        TreeNode *pCur=root;
        TreeNode *pPre=NULL;
        while(!sTree.empty()||pCur)
        {
            while(pCur!=NULL)
            {
                sTree.push(pCur);
                pCur=pCur->left;
            }
            pCur=sTree.top();
            if(pCur->right==NULL||pCur->right==pPre)
            {
                result.push_back(pCur->val);
                pPre=pCur;
                sTree.pop();
                pCur=NULL;
            }
            else
                pCur=pCur->right;
        }
        return result;
    }
};

 

方向思维：

 

/**
 * 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;
        if(root==NULL)
            return result;
        stack<TreeNode*> pTree;
        pTree.push(root);
        while(!pTree.empty())
        {
            TreeNode *curNode=pTree.top();
            result.push_back(curNode->val);
            pTree.pop();
            if(curNode->left)
                pTree.push(curNode->left);
            if(curNode->right)
                pTree.push(curNode->right);
        }
        reverse(result.begin(),result.end());
        return result;
    }
};