[Leetcode 94] 72 Binary Tree Inorder Traversal

Problem:

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

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

   1
    \
     2
    /
   3

 

return [1,3,2].

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

 

Analysis:

Basic tree traversal problems. Recursive solution is just to proper arrange the visiting order of the tree. Iterative version need the help of stack. My own solution is that everytime pop a node from the stack, if it has no children nodes, then record it in the result vector. Else we must first push it right child node,  then itself and at last its left child node. If any of the node is NULL, skip the pushing operation. Another thing to do is set the node's left and right pointer to NULL.

 

Code:

Recursive Version:

/**
 * 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> res;

    vector<int> inorderTraversal(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        res.clear();
        
        inorder(root);
        
        return res;
    }
    
    
private:
    void inorder(TreeNode *node) {
        if (node == NULL)
            return ;
            
        inorder(node->left);
        res.push_back(node->val);
        inorder(node->right);
    }
};

 

Iterative Version:

/**
 * 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> inorderTraversal(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> res;
        
        if (root == NULL)
            return res;
        
        stack<TreeNode *> stk;
        TreeNode *tmp, *node;
        stk.push(root);
        do {
            node = stk.top();
            stk.pop();
            
            if (node->left == NULL && node->right == NULL) {
                res.push_back(node->val);
                continue;
            }
            
            if (node->right != NULL) {
                stk.push(node->right);
                node->right = NULL;
            }
            
            if (node->left != NULL) {
                tmp = node->left;
                node->left = NULL;
                stk.push(node);
                stk.push(tmp);
            } else 
                stk.push(node);
            
        } while (!stk.empty());
        
        
        return res;
    }
};