Binary Tree Traversal 小结

本文总结了tree的三种traversal方式, 三种都用到stack。而且只有在inorder的时候while condition有所不同

1. Inorder Traversal
   https://leetcode.com/problems/binary-tree-inorder-traversal/

while(cur || !st.empty())， 同时while loop之前不 push stack

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ret;
        if(!root){
            return ret;
        }
        
        stack<TreeNode*> st;
        TreeNode *cur = root;
        
        while(cur || !st.empty()){
            if(cur){
                st.push(cur);
                cur = cur->left;
            }
            else{
                TreeNode *node = st.top(); st.pop();
                ret.push_back(node->val);
                cur = node->right;
            }
        }
        
        return ret;
    }
};

2. Preorder Traversal
   https://leetcode.com/problems/binary-tree-preorder-traversal/

while(!st.empty()), ， 同时while loop之前 push stack

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        if(!root){
            return vector<int>();
        }
        
        vector<int> ret;
        stack<TreeNode*> st;
        st.push(root);
        
        while(!st.empty()){
            TreeNode *node = st.top(); st.pop();
            ret.push_back(node->val);
            if(node->right){
                st.push(node->right);
            }
            
            if(node->left){
                st.push(node->left);
            }
        }
        
        return ret;
    }
};

3. Postorder Traversal:
   https://leetcode.com/problems/binary-tree-postorder-traversal/

Postorder 与 Preorder模板差不多，只是最后要reverse

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