binary-tree-postorder-traversal——二叉树后续遍历

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?

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:
     vector<int> postorderTraversal(TreeNode *root) {
         if(root==NULL)
             return v;
         TreeNode *left=root->left,*right=root->right;
         postorderTraversal(left);
         postorderTraversal(right);
         v.push_back(root->val);
         return v;
     }

     vector<int> v;
 };

非递归

处理并出栈时判断，该节点是否有子节点或其子节点是否被访问过。并记录上次处理的节点。

 /**
  * 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> v;
         if(root==NULL)
             return v;
         stack<TreeNode *> tree;
         tree.push(root);
         TreeNode *pre=NULL;
         while(!tree.empty()){
             TreeNode *p = tree.top();
             if((p->left==NULL&&p->right==NULL)||(pre!=NULL&&(p->left==pre||p->right==pre))){
                 v.push_back(p->val);
                 pre=p;
                 tree.pop();
             }else{
                 if(p->right!=NULL)
                     tree.push(p->right);
                 if(p->left!=NULL)
                     tree.push(p->left);
             }
         }
         return v;
     }
 };