LeetCode 94 Binary Tree Inorder Traversal (中序遍历二叉树)

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

For example:

Given binary tree [1,null,2,3],

   1
         2
    /
   3

return [1,3,2].

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

题目链接：https://leetcode.com/problems/binary-tree-inorder-traversal/

题目分析：中序遍历二叉树，递归的方法简单明了，不多说了

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    void DFS(TreeNode root, List<Integer> ans) {
        if(root == null) {
            return;
        }
        DFS(root.left, ans);
        ans.add(root.val);
        DFS(root.right, ans);
    }

    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        DFS(root, ans);
        return ans;
    }
}

下面是非递归的做法，用一个栈模拟递归的过程，中序遍历的顺序是 左->中->右，先一直走到最左端，取值后pop，这个过程其实完成了左->中的遍历，每次pop后再往右走，这样就完成了左->中->右的遍历

/**
 * 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> ans = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        while(root != null || !stack.empty()) {
            while(root != null) {
                stack.push(root);
                root = root.left;
            }
            if(!stack.empty()) {
                root = stack.peek();
                stack.pop();
                ans.add(root.val);
                root = root.right;
            }
        }
        return ans;
    }
}