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94. Binary Tree Inorder Traversal

时间:2018-11-11 23:27:12      阅读:163      评论:0      收藏:0      [点我收藏+]

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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?

 

Approach #1: recurisive.

/**
 * Definition for a binary tree node.
 * 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> ans;
        helper(root, ans);
        return ans;
    }
    
private:
    void helper(TreeNode* root, vector<int>& ans) {
        if (root != NULL) {
            if (root->left != NULL) {
                helper(root->left, ans);
            }
            ans.push_back(root->val);
            if (root->right != NULL) {
                helper(root->right, ans);
            }
        }
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Binary Tree Inorder Traversal.

 

Approach #2: Iteration with stack.[Java]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res =  new ArrayList< > ();
        Stack<TreeNode> stack = new Stack< > ();
        TreeNode curr = root;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            res.add(curr.val);
            curr = curr.right;
        }
        return res;
    }
}
Runtime: 1 ms, faster than 60.30% of Java online submissions for Binary Tree Inorder Traversal.

 

Approach #3: Morris Traversal.[python] 

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        res = []
        curr = root
        while curr:
            if not curr.left:
                res.append(curr.val)
                curr = curr.right
            else:
                pre = curr.left
                while pre.right:
                    pre = pre.right
                pre.right = curr
                temp = curr
                curr = curr.left
                temp.left = None
        return res            
            

Runtime: 24 ms, faster than 40.09% of Python online submissions for Binary Tree Inorder Traversal.

 

 

94. Binary Tree Inorder Traversal

标签:make   sam   uri   tor   else   sub   tput   comm   while   

原文地址:https://www.cnblogs.com/ruruozhenhao/p/9943581.html

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