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144. 二叉树的前序遍历

English Version

题目描述

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。

 

示例 1：

输入：root = [1,null,2,3]
输出：[1,2,3]

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：[1,2]

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

方法一：递归遍历

我们先访问根节点，然后递归左子树和右子树。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数，空间复杂度主要取决于递归调用的栈空间。

# 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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        def dfs(root):
            if root is None:
                return
            ans.append(root.val)
            dfs(root.left)
            dfs(root.right)

        ans = []
        dfs(root)
        return ans
/**
 * 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 {
    private List<Integer> ans = new ArrayList<>();

    public List<Integer> preorderTraversal(TreeNode root) {
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        ans.add(root.val);
        dfs(root.left);
        dfs(root.right);
    }
}
/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return;
            }
            ans.push_back(root->val);
            dfs(root->left);
            dfs(root->right);
        };
        dfs(root);
        return ans;
    }
};
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(root *TreeNode) (ans []int) {
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		ans = append(ans, root.Val)
		dfs(root.Left)
		dfs(root.Right)
	}
	dfs(root)
	return
}
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function preorderTraversal(root: TreeNode | null): number[] {
    const ans: number[] = [];
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return;
        }
        ans.push(root.val);
        dfs(root.left);
        dfs(root.right);
    };
    dfs(root);
    return ans;
}
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, ans: &mut Vec<i32>) {
        if root.is_none() {
            return;
        }
        let node = root.as_ref().unwrap().borrow();
        ans.push(node.val);
        Self::dfs(&node.left, ans);
        Self::dfs(&node.right, ans);
    }

    pub fn preorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut ans = vec![];
        Self::dfs(&root, &mut ans);
        ans
    }
}

方法二：栈实现非递归遍历

栈实现非递归遍历的思路如下：

  1. 定义一个栈 $stk$，先将根节点压入栈
  2. 若栈不为空，每次从栈中弹出一个节点
  3. 处理该节点
  4. 先把节点右孩子压入栈，接着把节点左孩子压入栈（如果有孩子节点）
  5. 重复 2-4
  6. 返回结果

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数，空间复杂度主要取决于栈空间。

# 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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        ans = []
        if root is None:
            return ans
        stk = [root]
        while stk:
            node = stk.pop()
            ans.append(node.val)
            if node.right:
                stk.append(node.right)
            if node.left:
                stk.append(node.left)
        return ans
/**
 * 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> preorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> stk = new ArrayDeque<>();
        stk.push(root);
        while (!stk.isEmpty()) {
            TreeNode node = stk.pop();
            ans.add(node.val);
            if (node.right != null) {
                stk.push(node.right);
            }
            if (node.left != null) {
                stk.push(node.left);
            }
        }
        return ans;
    }
}
/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        if (!root) {
            return ans;
        }
        stack<TreeNode*> stk;
        stk.push(root);
        while (stk.size()) {
            auto node = stk.top();
            stk.pop();
            ans.push_back(node->val);
            if (node->right) {
                stk.push(node->right);
            }
            if (node->left) {
                stk.push(node->left);
            }
        }
        return ans;
    }
};
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(root *TreeNode) (ans []int) {
	if root == nil {
		return
	}
	stk := []*TreeNode{root}
	for len(stk) > 0 {
		node := stk[len(stk)-1]
		stk = stk[:len(stk)-1]
		ans = append(ans, node.Val)
		if node.Right != nil {
			stk = append(stk, node.Right)
		}
		if node.Left != nil {
			stk = append(stk, node.Left)
		}
	}
	return
}
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function preorderTraversal(root: TreeNode | null): number[] {
    const ans: number[] = [];
    if (!root) {
        return ans;
    }
    const stk: TreeNode[] = [root];
    while (stk.length) {
        const { left, right, val } = stk.pop();
        ans.push(val);
        right && stk.push(right);
        left && stk.push(left);
    }
    return ans;
}

方法三：Morris 实现前序遍历

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 $ans$ 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，将当前节点值添加至结果列表 $ans$ 中，然后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。

时间复杂度 $O(n)$，其中 $n$ 是二叉树的节点数。空间复杂度 $O(1)$。

# 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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        ans = []
        while root:
            if root.left is None:
                ans.append(root.val)
                root = root.right
            else:
                prev = root.left
                while prev.right and prev.right != root:
                    prev = prev.right
                if prev.right is None:
                    ans.append(root.val)
                    prev.right = root
                    root = root.left
                else:
                    prev.right = None
                    root = root.right
        return ans
/**
 * 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> preorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        while (root != null) {
            if (root.left == null) {
                ans.add(root.val);
                root = root.right;
            } else {
                TreeNode prev = root.left;
                while (prev.right != null && prev.right != root) {
                    prev = prev.right;
                }
                if (prev.right == null) {
                    ans.add(root.val);
                    prev.right = root;
                    root = root.left;
                } else {
                    prev.right = null;
                    root = root.right;
                }
            }
        }
        return ans;
    }
}
/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        while (root) {
            if (!root->left) {
                ans.push_back(root->val);
                root = root->right;
            } else {
                TreeNode* prev = root->left;
                while (prev->right && prev->right != root) {
                    prev = prev->right;
                }
                if (!prev->right) {
                    ans.push_back(root->val);
                    prev->right = root;
                    root = root->left;
                } else {
                    prev->right = nullptr;
                    root = root->right;
                }
            }
        }
        return ans;
    }
};
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(root *TreeNode) (ans []int) {
	for root != nil {
		if root.Left == nil {
			ans = append(ans, root.Val)
			root = root.Right
		} else {
			prev := root.Left
			for prev.Right != nil && prev.Right != root {
				prev = prev.Right
			}
			if prev.Right == nil {
				ans = append(ans, root.Val)
				prev.Right = root
				root = root.Left
			} else {
				prev.Right = nil
				root = root.Right
			}
		}
	}
	return
}
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function preorderTraversal(root: TreeNode | null): number[] {
    const ans: number[] = [];
    while (root) {
        const { left, right, val } = root;
        if (!left) {
            ans.push(val);
            root = right;
        } else {
            let prev = left;
            while (prev.right && prev.right != root) {
                prev = prev.right;
            }
            if (!prev.right) {
                ans.push(val);
                prev.right = root;
                root = root.left;
            } else {
                prev.right = null;
                root = root.right;
            }
        }
    }
    return ans;
}