94. 二叉树的中序遍历

English Version

题目描述

给定一个二叉树的根节点 root ，返回它的 中序 遍历。

 

示例 1：

输入：root = [1,null,2,3]
输出：[1,3,2]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

示例 4：

输入：root = [1,2]
输出：[2,1]

示例 5：

输入：root = [1,null,2]
输出：[1,2]

 

提示：

• 树中节点数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

 

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

解法

1. 递归遍历

先递归左子树，再访问根节点，接着递归右子树。

2. 栈实现非递归遍历

非递归的思路如下：

1. 定义一个栈
2. 将树的左节点依次入栈
3. 左节点为空时，弹出栈顶元素并处理
4. 重复 2-3 的操作

3. Morris 实现中序遍历

Morris 遍历无需使用栈，空间复杂度为 O(1)。核心思想是：

遍历二叉树节点，

1. 若当前节点 root 的左子树为空，将当前节点值添加至结果列表 res 中，并将当前节点更新为 root.right
2. 若当前节点 root 的左子树不为空，找到左子树的最右节点 pre（也即是 root 节点在中序遍历下的前驱节点）：
   • 若前驱节点 pre 的右子树为空，将前驱节点的右子树指向当前节点 root，并将当前节点更新为 root.left。
   • 若前驱节点 pre 的右子树不为空，将当前节点值添加至结果列表 res 中，然后将前驱节点右子树指向空（即解除 pre 与 root 的指向关系），并将当前节点更新为 root.right。
3. 循环以上步骤，直至二叉树节点为空，遍历结束。

Python3

递归：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []

        def inorder(root):
            if root:
                inorder(root.left)
                res.append(root.val)
                inorder(root.right)

        inorder(root)
        return res

栈实现非递归：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res, s = [], []
        while root or s:
            if root:
                s.append(root)
                root = root.left
            else:
                root = s.pop()
                res.append(root.val)
                root = root.right
        return res

Morris 遍历:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        res = []
        while root:
            if root.left is None:
                res.append(root.val)
                root = root.right
            else:
                pre = root.left
                while pre.right and pre.right != root:
                    pre = pre.right
                if pre.right is None:
                    pre.right = root
                    root = root.left
                else:
                    res.append(root.val)
                    pre.right = None
                    root = root.right
        return res

Java

递归：

/**
 * 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> res = new ArrayList<>();
        inorder(root, res);
        return res;
    }

    private void inorder(TreeNode root, List<Integer> res) {
        if (root != null) {
            inorder(root.left, res);
            res.add(root.val);
            inorder(root.right, res);
        }
    }
}

栈实现非递归：

/**
 * 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> res = new ArrayList<>();
        Deque<TreeNode> s = new LinkedList<>();
        while (root != null || !s.isEmpty()) {
            if (root != null) {
                s.offerLast(root);
                root = root.left;
            } else {
                root = s.pollLast();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}

Morris 遍历:

/**
 * 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> res = new ArrayList<>();
        while (root != null) {
            if (root.left == null) {
                res.add(root.val);
                root = root.right;
            } else {
                TreeNode pre = root.left;
                while (pre.right != null && pre.right != root) {
                    pre = pre.right;
                }
                if (pre.right == null) {
                    pre.right = root;
                    root = root.left;
                } else {
                    res.add(root.val);
                    pre.right = null;
                    root = root.right;
                }
            }
        }
        return res;
    }
}

JavaScript

递归：

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var inorderTraversal = function (root) {
  let res = [];
  function inorder(root) {
    if (root) {
      inorder(root.left);
      res.push(root.val);
      inorder(root.right);
    }
  }
  inorder(root);
  return res;
};

栈实现非递归：

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var inorderTraversal = function (root) {
  let res = [];
  let s = [];
  while (root || s.length > 0) {
    if (root) {
      s.push(root);
      root = root.left;
    } else {
      root = s.pop();
      res.push(root.val);
      root = root.right;
    }
  }
  return res;
};

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        while (root)
        {
            if (root->left == nullptr)
            {
                res.push_back(root->val);
                root = root->right;
            } else {
                TreeNode* pre = root->left;
                while (pre->right && pre->right != root)
                {
                    pre = pre->right;
                }
                if (pre->right == nullptr)
                {
                    pre->right = root;
                    root = root->left;
                }
                else
                {
                    res.push_back(root->val);
                    pre->right = nullptr;
                    root = root->right;
                }
            }
        }
        return res;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func inorderTraversal(root *TreeNode) []int {
	var res []int
	for root != nil {
		if root.Left == nil {
			res = append(res, root.Val)
			root = root.Right
		} else {
			pre := root.Left
			for pre.Right != nil && pre.Right != root {
				pre = pre.Right
			}
			if pre.Right == nil {
				pre.Right = root
				root = root.Left
			} else {
				res = append(res, root.Val)
				pre.Right = nil
				root = root.Right
			}
		}
	}
	return res
}

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