LeetCode 94. 二叉树的中序遍历

题目

LeetCode 94. 二叉树的中序遍历

思路

中序遍历是按照左-根-右的顺序.

1. 首先要找到左-左-左….的那个节点
2. 打印后, 再判断它有没有右孩子, 若存在则继续迭代, 若无则打印当前节点
3. 打印完毕再从栈中取出之前存入的节点

代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new LinkedList<>();
        if(root == null) {
            return list;
        }
        Stack<TreeNode> stack = new Stack<>();
        while (!stack.isEmpty() || root != null) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            list.add(root.val);
            root = root.right;
        }
        return list;
    }
}