94. Binary Tree Inorder Traversal（非递归实现二叉树的中序遍历）

Posted 2021-02-15 shaer

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

Example:

Input: [1,null,2,3]
   1
         2
    /
   3

Output: [1,3,2]

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

 

方法一：递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public void preorderTraversal(TreeNode root) {
        if(root==null) return ;
        preorderTraversal(root.left);
        System.out.print(root.val+‘ ‘);
        preorderTraversal(root.right);
    }
}

 

方法二：迭代

中序遍历第左右根，所以设计程序时首先要考虑的是找到最左边的叶子结点。找到之后弹出还要考虑这个结点有没有右孩子。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        Stack<TreeNode> stack=new Stack<TreeNode>();
        List<Integer> list=new ArrayList<Integer>();
        while (root!=null||!stack.isEmpty()){
            while (root!=null){
                stack.add(root);
                root=root.left;
            }
            TreeNode treeNode=stack.pop();
            list.add(treeNode.val);
            root=treeNode.right;  //root是判断条件。每次弹出的结点都要检查是否还有右孩子。有就加入，没有就弹出。
        }
        return list;
    }
}