LeetCode--Binary Tree Postorder Traversal（栈实现三种遍历）

题意：Given a binary tree, return the postorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

题解：利用递归可轻易实现，要求用迭代，其实就是用自己写的栈来实现递归，因为DFS的递归实现本来就是用的栈。

DFS：

class Solution {
public:
    int dfs(TreeNode* now,vector<int>& ans){
        if(now == NULL)
            return 1;
        dfs(now->left,ans);
        dfs(now->right,ans);
        ans.push_back(now->val);
        return 1;
    }
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> ans;
        dfs(root,ans);
        return ans;
    }
};

迭代实现：

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> ans;
        if(root == NULL)
            return ans;
        stack<TreeNode*> sta;
        TreeNode* tmp;
        TreeNode* pre = new TreeNode(0);
        //上一个退栈的点，不能赋值NULL
        sta.push(root);
        while(!sta.empty()){
            tmp = sta.top();
            //模仿DFS，只可能有三种情况进入这，就是要加入左节点，要加入右节点，此节点退栈
            if(pre != tmp->left&&pre != tmp->right&&tmp->left != NULL){
                //当上一个退栈的点不是左节点也不是右节点，则需要加入左节点
                sta.push(tmp->left);
            }
            else if(pre != tmp->right&&tmp->right != NULL){
                //当上一个退栈的点不是右节点，则加入右节点
                sta.push(tmp->right);
            }
            else{
                //此节点退栈
                sta.pop();
                ans.push_back(tmp->val);
                pre = tmp;
            }
        }
        return ans;
    }
};

更好的方法是，因为最多第三次遍历某个节点，当第一次到这个节点时，我们要进入它的左节点并且把它的左节点置为NULL，下次到这个节点时就会直接检查右节点。当两个子节点都为空时，pop出此节点并打印val。

代码如下：

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ans;
        if(root == NULL) return ans;
        TreeNode* now;
        stack<TreeNode*> s;
        s.push(root);
        while(!s.empty()) {
            now = s.top();
            if(now->left) {
                s.push(now->left);
                now->left = NULL;
            }
            else if(now->right) {
                s.push(now->right);
                now->right = NULL;
            }
            else {
                ans.push_back(now->val);
                s.pop();
            }
        }
        return ans;
    }
};

Binary Tree Inorder Traversal 代码如下：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ans;
        if(root == NULL) return ans;
        stack<TreeNode*> s;
        s.push(root);
        while(!s.empty()) {
            TreeNode* now = s.top();
            if(now->left) {
                s.push(now->left);
                now->left = NULL;
            }
            else {
            //检查完左节点然后再进入时直接打印并出栈，再检查右节点
                ans.push_back(now->val);
                s.pop();
                if(now->right) {
                    s.push(now->right);
                }
            }
        }
        return ans;
    }
};

Binary Tree Preorder Traversal 代码如下：

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> s;
        s.push(root);
        vector<int> ans;
        TreeNode* now;
        TreeNode* next;
        if(root) {
        //在入栈的同时打印节点val值
            ans.push_back(root->val);
        }
        else {
            return ans;
        }
        while(!s.empty()) {
            now = s.top();
            if(now->left) {
                ans.push_back((now->left)->val);
                s.push(now->left);
                now->left = NULL;
            }
            else if(now->right) {
                ans.push_back((now->right)->val);
                s.push(now->right);
                now->right = NULL;
            }
            else {
                s.pop();
            }
        }
        return ans;
    }
};