【码不停题3.17】二叉树的前序后序遍历【递归非递归】

前序

非递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <stack>
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> mystack;
        vector<int> ans;
        if (!root) {
            return ans;
        }
        TreeNode* current = root;
        while (current || !mystack.empty()) {
            if (!current) {
                current = mystack.top();
                mystack.pop();
            }
            ans.push_back(current->val);
            if (current->right) {
                mystack.push(current->right);
            }
            current = current->left;
        }
        return ans;
    }
};

// 更好的写法
#include <stack>
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> mystack;
        vector<int> ans;
        if (!root) {
            return ans;
        }
        mystack.push(root);
        while (!mystack.empty()) {
            TreeNode* current = mystack.top();
            mystack.pop();
            ans.push_back(current->val);
            // 先右后左
            if (current->right) mystack.push(current->right);
            if (current->left) mystack.push(current->left);
        }
        return ans;
    }
};

递归

class Solution {
public:
    vector<int> ans;
    vector<int> preorderTraversal(TreeNode* root) {
        preorder(root);
        return ans;
    }
    void preorder(TreeNode* root) {
        if (root) {
            ans.push_back(root->val);
            preorder(root->left);
            preorder(root->right);
        }
    }
};

后序

非递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <stack>
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ans;
        stack<TreeNode*> mystack;
        if (!root) {
            return ans;
        }
        TreeNode* pre = nullptr;
        mystack.push(root);
        while (!mystack.empty()) {
            TreeNode* current = mystack.top();
            // 当前节点是叶子结点或者子节点都已被访问过
            if ((!current->left && !current->right) || (pre && (pre == current->left || pre == current->right)) ) {
                ans.push_back(current->val);
                pre = current;
                mystack.pop();
            } else {
            	// 先右后左
                if (current->right) mystack.push(current->right);
                if (current->left) mystack.push(current->left);
            }
        }
        return ans;
    }
};

// 由前序变换得来
#include <stack>
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> mystack;
        vector<int> ans;
        if (!root) {
            return ans;
        }
        mystack.push(root);
        while (!mystack.empty()) {
            TreeNode* current = mystack.top();
            mystack.pop();
            // 插在最前面
            ans.insert(ans.begin(), current->val);
            // 先左后右
            if (current->left) mystack.push(current->left);
            if (current->right) mystack.push(current->right);
        }
        return ans;
    }
};

递归

class Solution {
public:
    vector<int> ans;
    vector<int> postorderTraversal(TreeNode* root) {
        postorder(root);
        return ans;
    }
    void postorder(TreeNode* root) {
        if (root) {
            postorder(root->left);
            postorder(root->right);
            ans.push_back(root->val);
        }
    }
};

总结

1. 非递归 前序的每次都循环都有出栈 有出栈即有访问；后序的要满足要求才出栈，即满足要求才访问。
2. 前序 根 左 右 后序 左 右 根 。可以取巧用前序变形来替换后序。