LeetCode 94 Binary Tree Inorder Traversal(두 갈래 나무의 중서 반복)+(두 갈래 나무, 교체)

번역하다

 ， 。

 ：
  {1， #， 2， 3}
   1
    \
     2
    /
   3
  [1, 3, 2]

 ： ， ？

원문

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},
   1
    \
     2
    /
   3
return [1,3,2].

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

분석

/** * 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> v;

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

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> tempStack;
        while (!tempStack.empty() || root != NULL) {
            if (root != NULL) {
                tempStack.push(root);
                root = root->left;
            }
            else {
                root = tempStack.top();
                result.push_back((tempStack.top())->val);
                tempStack.pop();
                root = root->right;
            }
        }
        return result;
    }
};