[leet code] 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?

思路與 Binary Tree Preorder Traversal 相像. 如果使用遞歸, 則隻需要更改節點值提取順序.  順序改為: 左子節點值->節點值->右節點值

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> nodeVal = new ArrayList<Integer>();

        if (root == null) return nodeVal;
        
        TreeNode node = root;
        
        inOrder(nodeVal, node);
        
        return nodeVal;
    }
    
    public void inOrder(ArrayList<Integer> nodeVal, TreeNode node){
        if (node.left != null) inOrder(nodeVal, node.left);
        nodeVal.add(node.val);
        if (node.right != null) inOrder(nodeVal, node.right);
    }
}
           

如果不用遞歸, 參考 Binary Tree Preorder Traversal 可以利用棧進出實作inorder.  需要銘記在心的是, 對于每一個節點, 順序永遠是左子節點值->節點->右子節點值. 于是, 當檢驗任意節點時, 如果有左子節點, 則将目前節點壓入棧中, 并進而檢驗該節點左子樹, 不斷重複該操作直到棧清空且目前節點指向null.  這時可以提取儲存在棧中最頂端的節點 (該節點為中節點), 并檢查其右子樹. (該做法思路來自網上)

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> nodeVal = new ArrayList<Integer>();
        Stack<TreeNode> st = new Stack();
        
        if (root == null) return nodeVal;
        
        TreeNode node = root;
        
        while (!st.isEmpty() || node != null){
            //left node exist
            if (node != null){
                st.push(node);
                node = node.left;
            }
            else {
                // locate to the last unextract middle node
                node = st.peek();
                nodeVal.add(node.val);
                st.pop();
                node = node.right;
            }
        }
        return nodeVal;
    }
}