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},

return[3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

递归方法

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> postorderTraversal(TreeNode *root) {
13         if(root==NULL)
14             return v;
15         TreeNode *left=root->left,*right=root->right;
16         postorderTraversal(left);
17         postorderTraversal(right);
18         v.push_back(root->val);
19         return v;
20     }
21     
22     
23     vector<int> v;
24 };

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。

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> postorderTraversal(TreeNode *root) {
13         vector<int> v;
14         if(root==NULL)
15             return v;
16         stack<TreeNode *> tree;
17         tree.push(root);
18         TreeNode *pre=NULL;
19         while(!tree.empty()){
20             TreeNode *p = tree.top();
21             if((p->left==NULL&&p->right==NULL)||(pre!=NULL&&(p->left==pre||p->right==pre))){
22                 v.push_back(p->val);
23                 pre=p;
24                 tree.pop();
25             }else{
26                 if(p->right!=NULL)
27                     tree.push(p->right);
28                 if(p->left!=NULL)
29                     tree.push(p->left);
30             }
31         }
32         return v;
33     }
34 };