lintcode 访问路径
http://www.lintcode.com/zh-cn/problem/convert-bst-to-greater-tree/
描述
给出一个二叉搜索树，将每个节点重新赋值，本节点的值加上所有比本节点值大的节点的值。
样例
给出一个二叉搜索树 {5, 2, 3}：
返回一个新树
Java代码实现
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| /**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root the root of binary tree
* @return the new root
*/
public TreeNode convertBST(TreeNode root) {
// Write your code here
convert(root);
return root;
}
private int sum = 0;
private void convert(TreeNode root) {
if (root == null) {
return;
}
if (root.right != null) {
convert(root.right);
}
sum += root.val;
root.val = sum;
if (root.left != null) {
convert(root.left);
}
}
}
|
Python代码实现
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| """
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
# @param {TreeNode} root the root of binary tree
# @return {TreeNode} the new root
def convertBST(self, root):
# Write your code here
self.sum = 0
self.convert(root)
return root
def convert(self, root):
if root is None:
return
if root.right:
self.convert(root.right)
self.sum += root.val
root.val = self.sum
if root.left:
self.convert(root.left)
|