给定一个二叉树，返回它的 后序 遍历。

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

Solution

参考 @z1m 、[94 二叉树的中序遍历]、[144 二叉树的前序遍历]、[102 二叉树的层序遍历]

• 递归法

// @lc code=start
/**
 * 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> postorderTraversal(TreeNode* root) {
        vector<int> res;
        postorder(root, res);
        return res;
    }
private:
    void postorder(TreeNode* root, vector<int>& res){
        if(root){
            postorder(root->left, res);
            postorder(root->right, res);
            res.push_back(root->val);
        }
    }
};
// @lc code=end

• 迭代法

• 当各个节点以“左->右->中”的次序依次出现在迭代的loop当中时，它才是真正的后序遍历

• 根结点如果有右孩子，需要重新入栈

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stack;
        TreeNode* node = root, *pre=nullptr;
        while(node || !stack.empty()){
            while(node){
                stack.push(node);
                node = node->left;
            }
            node = stack.top();
            stack.pop();
            if(node->right==nullptr || node->right==pre){
                res.push_back(node->val);
                pre = node;
                node = nullptr;
            } else {
                stack.push(node);
                node = node->right;
            }
        }
        return res;
    }
};

• 方案二：访问过程 “根 -> 右 -> 左”，将结果反转
• 本质上，这种方法是一个“前序遍历”，而不是“后序遍历”。

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stack;
        TreeNode* node = root;
        while(node || !stack.empty()){
            while(node){
                res.push_back(node->val);
                stack.push(node);
                node = node->right;
            }
            node = stack.top();
            stack.pop();
            node = node->left;
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

java

• 递归法

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        postorder(root, result);
        return result;
    }

    private void postorder(TreeNode root, List<Integer> rec) {
        if (root == null) {
            return;
        }
        postorder(root.left, rec);
        postorder(root.right, rec);
        rec.add(root.val);
    }
}

• “伪”迭代法

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        Deque<TreeNode> stack = new LinkedList<>();
        TreeNode node = root;
        while (node != null || !stack.isEmpty()) {
            while (node != null) {
                result.add(node.val);
                stack.push(node);
                node = node.right;
            }
            node = stack.pop();
            node = node.left;
        }
        Collections.reverse(result);
        return result;
    }
}