题解 | 实现二叉树先序，中序和后序遍历

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 the root of binary tree
     * @return int整型vector<vector<>>
     */
    vector<int> pre;
    vector<int> mid;
    vector<int> post;

    vector<vector<int> > threeOrders(TreeNode* root) {
        if(root != nullptr){
            preorder(root);
            midorder(root);
            postorder(root);
        }

        vector<vector<int>> res = {pre, mid, post};
        return res;
    }

    void preorder(TreeNode* node){   //前序遍历：中左右；入栈右左，出栈即记录入数组并再入栈子节点；
        stack<TreeNode*> stk;
        stk.push(node);
        while(!stk.empty()){
            TreeNode* temp = stk.top();stk.pop();
            if(temp){
            pre.push_back(temp->val);

            stk.push(temp->right);
            stk.push(temp->left);
            }
        }
    }

    void midorder(TreeNode* node){  //中序遍历
        stack<TreeNode*> stk;    //没有先入根节点
        while(node || !stk.empty()){  //先把所有的左子节点压入栈中
            while(node){
                stk.push(node);
                node = node->left;
            }

            node = stk.top();stk.pop();  //到头了，就出最左边的，左右并压入其右节点
            mid.push_back(node->val);
            node  = node->right;
        }
    }

    void postorder(TreeNode* node){   //后序遍历：左右中；
        stack<TreeNode*> stk;
        stk.push(node);
        while(!stk.empty()){
            TreeNode* temp = stk.top();stk.pop();
            if(!temp)continue;
            post.push_back(temp->val);
            stk.push(temp->left);
            stk.push(temp->right);
        }
        reverse(post.begin(), post.end()); //入栈中，左右；最后进行反转；
    }
};