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Leetcode[94] Binary Tree Inorder Traversal

07 Nov 2015

###Task1 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?

###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 inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        ret = []
        self.inorder(ret, root)
        return ret
    def inorder(self, ret, root):
        if root == None:
            return
        self.inorder(ret, root.left)
        ret.append(root.val)
        self.inorder(ret, root.right)

####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 inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        ret = []
        stack = []
        cur = root
        while cur or len(stack) > 0:
            while cur:
                stack.append(cur)
                cur = cur.left
            l = stack.pop()
            ret.append(l.val)
            cur = l.right
        return ret

###Easy to understand

# 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]
        """
        ret = []
        stack = []
        cur = root
        while cur or len(stack) > 0:
            if cur:
                stack.append(cur)
                cur = cur.left
            elif len(stack) > 0:
                newRoot = stack.pop()
                ret.append(newRoot.val)
                cur = newRoot.right
        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> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        while (!stack.isEmpty() || cur != null) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left; 
            }
            TreeNode poped = stack.pop();
            cur = poped.right;
            res.add(poped.val);
        }
        
        return res;
    }
}

###Points

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