Thursday, February 5, 2015

LeetCode [94] Binary Tree Inorder Traversal

 94. Binary Tree Inorder Traversal

Medium

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

Example:

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

Output: [1,3,2]

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

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//C++
//C++: dfs 0ms
/**
 * 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) {
        vector<int> res;
        dfs(root, res);
        return res;
    }
    void dfs(TreeNode* root, vector<int> &res){
        if(!root) return;
        dfs(root->left, res);
        res.push_back(root->val);
        dfs(root->right, res);
    }
};
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stk;
        TreeNode *p = root;

        while(p||!stk.empty()){
            while(p){
                stk.push(p);
                p = p->left;
            }
            p = stk.top();
            stk.pop();
            res.push_back(p->val);
            p = p->right;
        }

        return res;
    }
};
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> res;
        TreeNode *c = root;
        while(c){
            if(c->left){
                TreeNode *p = c->left;
                while(p->right && p->right!=c){
                    p = p->right;
                }
                if(p->right==NULL){
                    p->right = c;
                    c = c->left;
                }else{//c == p->right;
                    p->right = NULL;
                    res.push_back(c->val);
                    c = c->right;
                }
            }else{
                res.push_back(c->val);
                c = c->right;
            }
        }
        return res;
    }
};

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//Java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    List<Integer> ret = new ArrayList();
    public List<Integer> inorderTraversal(TreeNode root) {
        helper(root);
        return ret;
    }
    
    void helper(TreeNode root){
        if(root!=null){
            helper(root.left);
            ret.add(root.val);
            helper(root.right);
        }
    }
}


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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        if(root==null) return new ArrayList<>();
        TreeNode p = root;
        List<Integer> list = new ArrayList<>();
        while(p!=null){
            if(p.left==null){
                list.add(p.val);
                p = p.right;
            }else{//p.left!=null
                TreeNode prev = p.left;
                while(prev.right!=null && prev.right!=p) prev = prev.right;
                if(prev.right==null){//first visit the left of p
                    prev.right = p;
                    p = p.left;
                }else{//second time visit the left of p => we've done with p.left
                    prev.right = null;
                    list.add(p.val);
                    p = p.right;
                }
            }
        }
        
        return list;
    }
}
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