LeetCode(50) Binary Tree Inorder Traversal 中序遍历

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题目描述

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

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

return [1,3,2].

解题思路

之前在二叉树的先序遍历中已经介绍了二叉树遍历的思路了，这里直接上代码。

递归解法

/**
 * 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;
    }
};

非递归解法

/**
 * 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;
    }
};