題目描述

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

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

解題思路

之前在二叉樹的先序遍歷中已經介紹了二叉樹遍歷的思路了，這裡直接上程式碼。

遞迴解法

/**
 * 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:
    void traversal(TreeNode* node, vector<int>& node_value)
    {
        if(!node) return;
        traversal(node->left, node_value);
        node_value.push_back(node->val);
        traversal(node->right, node_value);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector
 
<int> node_value;
        traversal(root, node_value);
        return node_value;
    }
};

非遞迴解法

/**
 * 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> node_value;
        if(!root) return node_value;

        stack<TreeNode*> nodes;
        while(root)
        {
            nodes.push(root);
            root = root->left;
        }

        while(!nodes.empty())
        {
            TreeNode* top = nodes.top();
            node_value.push_back(top->val);
            nodes.pop();
            top = top->right;
            while(top)
            {
                nodes.push(top);
                top = top->left;
            }
        }

        return node_value;
    }
};