Binary Tree Inorder Traversal(转)

 

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].

 

/**
 * 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<>();
        dfs(root, res);
        return res;
    }

    private void dfs(TreeNode root, List<Integer> res) {
        if (root != null) {
            dfs(root.left, res);
            res.add(root.val);
            dfs(root.right, res);
        }
    }
}

 

/**
 * 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<>();
        if (root == null) {
            return res;
        }
        LinkedList<TreeNode> stack = new LinkedList<>();
        while (root!=null || !stack.isEmpty()) {
            if (root!=null) {
                stack.push(root);
                root = root.left;
            } else {
                root = stack.pop();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}

http://hcx2013.iteye.com/blog/2230218