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Leetcode[144] Binary Tree Preorder Traversal

11 Nov 2015

###Task1 Given a binary tree, return the preorder traversal of its nodes’ values.

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

1

\ 
 2 
/

3

return [1,2,3].

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

###Python ####Recursive

# 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 preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        ret = []
        self.preorder(root, ret)
        return ret
    def preorder(self, root, ret):
        if not root:
            return
        ret.append(root.val)
        self.preorder(root.left, ret)
        self.preorder(root.right, ret)

####Iterative

# 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 preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        stack = []
        ret = []
        stack.append(root)
        while len(stack) > 0:
            cur = stack.pop()
            ret.append(cur.val)
            if cur.right:
                stack.append(cur.right)
            if cur.left:
                stack.append(cur.left)
        return ret

###Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            if (cur.right != null) {
                stack.push(cur.right);
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }
            res.add(cur.val);
        }
        
        return res;
    }
}

###Points

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