[leetcode]Binary Tree InorderTraversal

Binary Tree Inorder Traversal

 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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

算法思路：

1. 递归实现

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    List<Integer> res = new ArrayList<Integer>();
    public List<Integer> inorderTraversal(TreeNode root) {
        if(root == null) return res;
        if(root.left != null) inorderTraversal(root.left);
        res.add(root.val);
        if(root.right != null) inorderTraversal(root.right);
        return res;
    }
}

 

思路2：

非递归版本：

借用栈，将当前节点的所有左节点（左节点的左节点....）压栈，然后依次弹栈处理当前节点并处理右子树

代码如下：

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 current = root;
        while(true){
            while(current != null){
                stack.push(current);
                current = current.left;
            }
            if(stack.isEmpty()) break;
            current = stack.pop();
            res.add(current.val);
            current = current.right;
        }
        return res;
    }
}