LeetCode 145. Binary Tree Postorder Traversal

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].

Stack Problem, One example will make this question easy to tackle.

In this example, the post-order sequence is: 4, 5, 2, 6, 3, 1.  

The process will be.... We first push 1 onto stack, Get the top element, pop it off, then push the left and right child.....

vector<int> postorderTraversal(TreeNode* root) {        if(!root) return {};        vector<int> res;        stack<TreeNode*> nodes;        nodes.push(root);        while(!nodes.empty()) {            TreeNode* tmp = nodes.top();            nodes.pop();            res.push_back(tmp->val);            if(tmp->left) nodes.push(tmp->left);            if(tmp->right) nodes.push(tmp->right);        }        reverse(res.begin(), res.end());        return res;    }

Another way of asking this question is to implement a post-order iterator which has getNext() method, and hasNext() method.

   1  / \ 4   2    / \   3   6   The post order is: 4 3 6 2 1 (left substree, root, right substree)

class postOrderIterator {  stack<TreeNode*> nodes;public:  void findNextLeave(TreeNode* root) {    TreeNode* curr = root;    while(curr) {      nodes.push_back(curr);      if(curr->left) curr = curr->left;      else if(curr->right) curr = curr->right;    }  }  bool hasNext() {    return !nodes.empty();  }  int getNext() {    TreeNode* res = nodes.top();    nodes.pop();    while(!nodes.empty()) {      TreeNode* tmp = nodes.top();      if(res == tmp->left) {  // left is done, traversal the right.        findNextLeave(tmp->right);      }    }    return res->val;  }}