[LeetCode] Binary Tree Inorder Traversal

Binary Tree Inorder Traversal My Submissions Question Solution Total
Accepted: 85535 Total Submissions: 232962 Difficulty: Medium 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?

confused what “{1,#,2,3}” means? > read more on how binary tree is
serialized on OJ.

二叉树的中序遍历

递归的方法

/**
 * 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> inorderTraversal(TreeNode* root) {
        vector<int> v;
        recur_inorder(root,v);
        return v;
    }

    void recur_inorder(TreeNode* T, vector<int> &v){
        if(T!=NULL){
            if(T->left!=NULL)
                recur_inorder(T->left,v);
            v.push_back(T->val);
            if(T->right!=NULL)
                recur_inorder(T->right,v);
        }
    }
};

非递归使用stack

/**
 * 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> inorderTraversal(TreeNode* root) {
        stack<TreeNode*> sta;
        vector<int> v;
        TreeNode* T= root;
        // sta.push(T);
        while(!sta.empty()||T){
            while(T!=NULL){
                sta.push(T);
                 T= T->left;
            }
            T= sta.top();
            sta.pop();
            v.push_back(T->val);
            T = T->right;
        }
        return v;
    }
};

Morris Traversal方法遍历二叉树（非递归，不用栈，O(1)空间）

O(1)space的方法

这里要重点讲的方法