LeetCode145. Binary Tree Postorder Traversal(后序遍历)

145. Binary Tree Postorder Traversal

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

Example:

Input: [1,null,2,3]

   1
    \
     2
    /
   3

Output: [3,2,1]

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

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题目：二叉树的后续遍历，递归的方式。

思路：Preorder, Inorder, and Postorder Iteratively Summarization

工程代码下载

/**
 * 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) {
        stack<TreeNode*> nodes;
        vector<int> res;
        while(root != nullptr || nodes.size()){
            if(root != nullptr){
                nodes.push(root);
                res.push_back(root->val);
                root = root->right;
            }
            else{
                TreeNode* node = nodes.top();
                nodes.pop();
                root = node->left;
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};