leetcode Binary Tree Postorder Traversal-优快云博客

本文介绍三种不同的二叉树后序遍历迭代方法：模拟递归程序、使用两个栈及记录前一节点访问状态。这些方法避免了递归带来的栈溢出风险，适用于需要高效遍历二叉树的场景。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [3,2,1].

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

Solution1 : Simulate the recursive program:

Note that: 1. sta.pop();

      if (!sta.empty()) {

not

      while (!sta.empty()) {

The right code is :

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
 public:
  vector<int> postorderTraversal(TreeNode *root) {
    stack<pair<TreeNode*, int>> sta;
    vector<int> res;
    while (root != NULL || !sta.empty()) {
      while (root != NULL) {
        sta.push(make_pair(root, 0)); //0 stands for right child and itself;
        root = root->left;
      }    
      if (!sta.empty()) {
        root = sta.top().first;
        int status = sta.top().second;
        sta.pop();
        if (status == 0) {
          sta.push(make_pair(root, 1));
          root = root->right;
        }
        else if (status == 1){
          res.push_back(root->val);
          root = NULL;
        }
      }
    }
    return res;
  }
};

Solution 2: Using two stacks:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
 public:
  vector<int> postorderTraversal(TreeNode *root) {
    stack<TreeNode*> s,output;
    vector<int> res;
    TreeNode* cur = NULL;
    if (root != NULL)
      s.push(root);
    while (!s.empty()) {
      cur = s.top();
      s.pop();
      output.push(cur);
      if (cur->left)
        s.push(cur->left);
      if (cur->right)
        s.push(cur->right);
    }
    while (!output.empty()) {
      res.push_back(output.top()->val);
      output.pop();
    }
    return res;
  }
};

Solution 3: Record the previous node, when cur->left == prev, push the right child to stack for the next-turn visiting:

A wrong code:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
 public:
  vector<int> postorderTraversal(TreeNode *root) {
    vector<int> res;
    TreeNode *prev = NULL, *cur = NULL;
    stack<TreeNode*> s;
    if (root != NULL)
      s.push(root);
    while (!s.empty()) {
      cur = s.top();
      if (prev == NULL || prev->left == cur || prev->right == cur) 
        if (cur->left)
          s.push(cur->left);
        else if (cur->right)
          s.push(cur->right);
      else if (cur->left == prev && cur->right)
        s.push(cur->right);
      else {
        res.push_back(cur->val);
        s.pop();
      }
      prev = cur;
    }
    return res;
  }
};

The right code is :

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
 public:
  vector<int> postorderTraversal(TreeNode *root) {
    vector<int> res;
    TreeNode *prev = NULL, *cur = NULL;
    stack<TreeNode*> s;
    if (root != NULL)
      s.push(root);
    while (!s.empty()) {
      cur = s.top();
      if (prev == NULL || prev->left == cur || prev->right == cur) {
        if (cur->left)
          s.push(cur->left);
        else if (cur->right)
          s.push(cur->right);
      }
      else if (cur->left == prev && cur->right)
        s.push(cur->right);
      else {
        res.push_back(cur->val);
        s.pop();
      }
      prev = cur;
    }
    return res;
  }
};