题目描述:

Given a binary tree, return the postorder traversal of its nodes’ values.

For example: Given binary tree {1,#,2,3},

1
2
3
4
5
6
1
 \
  2
 /
3

return [3,2,1].

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

非递归后序遍历.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
/**
 * Definition for a binary tree node.
 * 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> ans;
        if(!root) return ans;
        vector<TreeNode*> stack = {root};
        
        while(!stack.empty()){
            TreeNode* n = stack.back();
            stack.pop_back();
            ans.push_back(n->val);
            if(n->left) stack.push_back(n->left);
            if(n->right) stack.push_back(n->right);
        }
        reverse(ans.begin(), ans.end());
        
        return ans;
    }
};