leetcode之binary-tree-inorder-traversal（二叉樹中序周遊）

題目

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

For example:

Given binary tree{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.

題意

給定一個二叉樹，求其i中序周遊的輸出；

C++實作代碼

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 //非遞歸版本，使用stack，先全部擷取所有左子樹節點，然後再逐級向上，求其右子樹
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
       stack<TreeNode *> s;
       vector<int> v;
       if(root ==nullptr)
           return v;
        TreeNode *node = root;
        while(node != nullptr || !s.empty()){
            while(node){
                s.push(node);
                node = node->left;
            }
            if(!s.empty()){
                node = s.top();
                s.pop();
                v.push_back(node->val);
                node = node->right;
            }
        }
        return v;
    }

};
           

C++遞歸版本

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> v;
        inorder(root,v);
        return v;
    }
    void inorder(TreeNode *root,vector<int> & v){
        if(root ==nullptr)
            return ;
        if(root->left){
            inorder(root->left,v);
        }
        v.push_back(root->val);
        if(root->right){
            inorder(root->right,v);
        }
    }
};