Leetcode dfs Binary Tree Postorder Traversal

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Binary Tree Postorder Traversal Total Accepted: 28560 Total Submissions: 92333My Submissions

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 iterativel

題意：後序遍曆，不過給出了函式宣告，限制了實現方式
思路：採用遞迴實現。因為函式宣告是返回一個vector<int>,所以每個子樹返回的是該子樹的後序遍曆的結果
按照 左、右、根的次序把根和左右子樹的vector合并起來就可以了
/**
 * 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> post;
    if(root == NULL) return post;
    TreeNode *left = root->left;
    TreeNode *right = root->right;
    
    if(left) {
    vector<int> left_vector = postorderTraversal(left);
    post.insert(post.end(), left_vector.begin(), left_vector.end());
    }
    if(right){
    vector<int> right_vector = postorderTraversal(right);
    post.insert(post.end(), right_vector.begin(), right_vector.end());
    }
    post.push_back(root->val);
    
    return post;
    }
};

Leetcode dfs Binary Tree Postorder Traversal