【二叉树系列 | 01】二叉树遍历

 前序遍历

 思路：前序遍历算法先访问树的根节点，然后遍历左子树，最后遍历右子树。

解法1

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) 
    {
        if(!root) return {};
        stack<TreeNode *> st;
        vector<int> res;
        TreeNode *p = root;
        st.push(p);
        while(!st.empty())
        {
            p = st.top();
            st.pop();
            res.push_back(p->val);
            
            if(p->right) st.push(p->right);
            if(p->left) st.push(p->left);
        }
        return res;
    }
};

中序遍历

思路：中序遍历的算法先遍历左子树，然后访问根节点，最后遍历右子树。

解法1：非递归遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) 
    {
        if(!root) return {};
        
        vector<int> res;
        stack<TreeNode *> st;
        TreeNode *p = root;
        
        while(!st.empty() || p)
        {
            while(p)
            {
                st.push(p);
                p = p->left;
            }
            p = st.top();
            st.pop();
            res.push_back(p->val);
            
            p = p->right;
        }
        return res;
    }
};