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中等
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第 135 场周赛 Q2
树
深度优先搜索
二叉搜索树
二叉树

1038. 从二叉搜索树到更大和树

English Version

题目描述

给定一个二叉搜索树 root (BST)，请将它的每个节点的值替换成树中大于或者等于该节点值的所有节点值之和。

提醒一下， 二叉搜索树 满足下列约束条件：

  • 节点的左子树仅包含键 小于 节点键的节点。
  • 节点的右子树仅包含键 大于 节点键的节点。
  • 左右子树也必须是二叉搜索树。

 

示例 1：

输入：[4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
输出：[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

示例 2：

输入：root = [0,null,1]
输出：[1,null,1]

 

提示：

  • 树中的节点数在 [1, 100] 范围内。
  • 0 <= Node.val <= 100
  • 树中的所有值均 不重复 。

 

注意：该题目与 538: https://leetcode.cn/problems/convert-bst-to-greater-tree/  相同

解法

方法一：递归

按照“右根左”的顺序，递归遍历二叉搜索树，累加遍历到的所有节点值到 $s$ 中，然后每次赋值给对应的 node 节点。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉搜索树的节点数。

Python3

# 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 bstToGst(self, root: TreeNode) -> TreeNode:
        def dfs(root):
            nonlocal s
            if root is None:
                return
            dfs(root.right)
            s += root.val
            root.val = s
            dfs(root.left)

        s = 0
        dfs(root)
        return root

Java

/**
 * 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 int s;

    public TreeNode bstToGst(TreeNode root) {
        dfs(root);
        return root;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.right);
        s += root.val;
        root.val = s;
        dfs(root.left);
    }
}

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:
    int s = 0;

    TreeNode* bstToGst(TreeNode* root) {
        dfs(root);
        return root;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        dfs(root->right);
        s += root->val;
        root->val = s;
        dfs(root->left);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func bstToGst(root *TreeNode) *TreeNode {
	s := 0
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Right)
		s += root.Val
		root.Val = s
		dfs(root.Left)
	}
	dfs(root)
	return root
}

TypeScript

/**
 * 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 bstToGst(root: TreeNode | null): TreeNode | null {
    const dfs = (root: TreeNode | null, sum: number) => {
        if (root == null) {
            return sum;
        }
        const { val, left, right } = root;
        sum = dfs(right, sum) + val;
        root.val = sum;
        sum = dfs(left, sum);
        return sum;
    };
    dfs(root, 0);
    return root;
}

Rust

// 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::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(root: &mut Option<Rc<RefCell<TreeNode>>>, mut sum: i32) -> i32 {
        if let Some(node) = root {
            let mut node = node.as_ref().borrow_mut();
            sum = Self::dfs(&mut node.right, sum) + node.val;
            node.val = sum;
            sum = Self::dfs(&mut node.left, sum);
        }
        sum
    }

    pub fn bst_to_gst(mut root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
        Self::dfs(&mut root, 0);
        root
    }
}

JavaScript

/**
 * 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 {TreeNode}
 */
var bstToGst = function (root) {
    let s = 0;
    function dfs(root) {
        if (!root) {
            return;
        }
        dfs(root.right);
        s += root.val;
        root.val = s;
        dfs(root.left);
    }
    dfs(root);
    return root;
};

C

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

int dfs(struct TreeNode* root, int sum) {
    if (root) {
        sum = dfs(root->right, sum) + root->val;
        root->val = sum;
        sum = dfs(root->left, sum);
    }
    return sum;
}

struct TreeNode* bstToGst(struct TreeNode* root) {
    dfs(root, 0);
    return root;
}

方法二：Morris 遍历

Morris 遍历无需使用栈，时间复杂度 $O(n)$，空间复杂度为 $O(1)$。核心思想是：

定义 s 表示二叉搜索树节点值累加和。遍历二叉树节点：

  1. 若当前节点 root 的右子树为空，将当前节点值添加至 s 中，更新当前节点值为 s，并将当前节点更新为 root.left。
  2. 若当前节点 root 的右子树不为空，找到右子树的最左节点 next（也即是 root 节点在中序遍历下的后继节点）：
    • 若后继节点 next 的左子树为空，将后继节点的左子树指向当前节点 root，并将当前节点更新为 root.right。
    • 若后继节点 next 的左子树不为空，将当前节点值添加 s 中，更新当前节点值为 s，然后将后继节点左子树指向空（即解除 next 与 root 的指向关系），并将当前节点更新为 root.left。
  3. 循环以上步骤，直至二叉树节点为空，遍历结束。
  4. 最后返回二叉搜索树根节点即可。

Morris 反序中序遍历跟 Morris 中序遍历思路一致，只是将中序遍历的“左根右”变为“右根左”。

Python3

# 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 bstToGst(self, root: TreeNode) -> TreeNode:
        s = 0
        node = root
        while root:
            if root.right is None:
                s += root.val
                root.val = s
                root = root.left
            else:
                next = root.right
                while next.left and next.left != root:
                    next = next.left
                if next.left is None:
                    next.left = root
                    root = root.right
                else:
                    s += root.val
                    root.val = s
                    next.left = None
                    root = root.left
        return node

Java

/**
 * 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 TreeNode bstToGst(TreeNode root) {
        int s = 0;
        TreeNode node = root;
        while (root != null) {
            if (root.right == null) {
                s += root.val;
                root.val = s;
                root = root.left;
            } else {
                TreeNode next = root.right;
                while (next.left != null && next.left != root) {
                    next = next.left;
                }
                if (next.left == null) {
                    next.left = root;
                    root = root.right;
                } else {
                    s += root.val;
                    root.val = s;
                    next.left = null;
                    root = root.left;
                }
            }
        }
        return node;
    }
}

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:
    TreeNode* bstToGst(TreeNode* root) {
        int s = 0;
        TreeNode* node = root;
        while (root) {
            if (root->right == nullptr) {
                s += root->val;
                root->val = s;
                root = root->left;
            } else {
                TreeNode* next = root->right;
                while (next->left && next->left != root) {
                    next = next->left;
                }
                if (next->left == nullptr) {
                    next->left = root;
                    root = root->right;
                } else {
                    s += root->val;
                    root->val = s;
                    next->left = nullptr;
                    root = root->left;
                }
            }
        }
        return node;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func bstToGst(root *TreeNode) *TreeNode {
	s := 0
	node := root
	for root != nil {
		if root.Right == nil {
			s += root.Val
			root.Val = s
			root = root.Left
		} else {
			next := root.Right
			for next.Left != nil && next.Left != root {
				next = next.Left
			}
			if next.Left == nil {
				next.Left = root
				root = root.Right
			} else {
				s += root.Val
				root.Val = s
				next.Left = nil
				root = root.Left
			}
		}
	}
	return node
}

TypeScript

/**
 * 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 bstToGst(root: TreeNode | null): TreeNode | null {
    let cur = root;
    let sum = 0;
    while (cur != null) {
        const { val, left, right } = cur;
        if (right == null) {
            sum += val;
            cur.val = sum;
            cur = left;
        } else {
            let next = right;
            while (next.left != null && next.left != cur) {
                next = next.left;
            }
            if (next.left == null) {
                next.left = cur;
                cur = right;
            } else {
                next.left = null;
                sum += val;
                cur.val = sum;
                cur = left;
            }
        }
    }
    return root;
}

C

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode* bstToGst(struct TreeNode* root) {
    struct TreeNode* cur = root;
    int sum = 0;
    while (cur) {
        if (!cur->right) {
            sum += cur->val;
            cur->val = sum;
            cur = cur->left;
        } else {
            struct TreeNode* next = cur->right;
            while (next->left && next->left != cur) {
                next = next->left;
            }
            if (!next->left) {
                next->left = cur;
                cur = cur->right;
            } else {
                next->left = NULL;
                sum += cur->val;
                cur->val = sum;
                cur = cur->left;
            }
        }
    }
    return root;
}