中序遍历定义

首先遍历左子树，然后访问根结点，最后遍历右子树。
遵循：左根右

如图所示：中序遍历结果为4 2 5 1 3

解释图：

leetcode 94.二叉树的中序遍历

法一：利用栈

  /**
 * 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) {
        vector<int> res;
        if(!root)
            return res;
        stack<TreeNode*> st;
        TreeNode* node = root;
        while(1){
            while(node){//进入左子树
                st.push(node);
                node = node->left;
            }
            if(st.empty())//若为空，则遍历完所有的结点
                break;
            node = st.top();
            st.pop();
            res.push_back(node->val);//将结点值放入数组
            node = node->right;//进入右子树
        }
        return res;
    }
};

法二：递归

/**
 * 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> res;//注意放在外面
    vector<int> inorderTraversal(TreeNode* root) {
        if(root!=NULL){
            inorderTraversal(root->left);//沿左子树向下走
            res.push_back(root->val);
            inorderTraversal(root->right);
        }
        return res;
    }
};