算法：中序遍历二叉树94. Binary Tree Inorder Traversal

Posted 2021-11-22 架构师易筋

94. Binary Tree Inorder Traversal

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

Example 1:

Input: root = [1,null,2,3]
Output: [1,3,2]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Example 4:

Input: root = [1,2]
Output: [2,1]

Example 5:

Input: root = [1,null,2]
Output: [1,2]

Constraints:

• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

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

递归算法

/**
 * 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 ArrayList<>();
        helper(root, list);
        
        return list;
    }
    
    private void helper(TreeNode root, List<Integer> list) {
        if (root == null) return;
        helper(root.left, list);
        list.add(root.val);
        helper(root.right, list);
    }
}

遍历解法

将一个 Node 的所有左子节点都压入堆栈，这样我的想法就很清楚了：

1. 我们将 root 的所有左孩子推入堆栈，直到没有更多节点。
2. 然后我们从我们调用的堆栈中弹出cur。
3. 添加cur到结果列表
4. 设置root = node.right， 以此循环。

/**
 * 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 ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        while (root != null || !stack.isEmpty()) {
            while(root != null) {
                stack.add(root);
                root = root.left;
            }
            TreeNode node = stack.pop();
            list.add(node.val);
            root = node.right;
        }
        
        return list;
    }
}