[Leetcode] Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

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

return [1,3,2].

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

[Thoughts]

先根遍历的迭代算法比较简单，但是中根遍历的迭代算法稍微难一点。同样借助于栈，但不同于先根遍历的中规中矩，中根遍历需要先把当前结点的所有左结点放入栈中，然后从栈中弹出前一个左结点(即当前的根结点)，把值放入队列，然后指向右结点，重复直至栈空。

[Code]

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

[Leetcode] Binary Tree Inorder Traversal

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