题目
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/ 2 2
/ \ / 3 4 4 3
...
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其他好文 时间:
2014-06-08 18:12:04
阅读次数:
248
题目
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
G...
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其他好文 时间:
2014-06-08 17:29:07
阅读次数:
315
1. 递归解法
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
cl...
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其他好文 时间:
2014-06-08 16:51:59
阅读次数:
199
【题目】
Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
You may only use constant extra space.
For example,
Given the following binary tre...
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其他好文 时间:
2014-06-08 15:46:22
阅读次数:
303
题目
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:
A solution using O(n)
space is pretty straight forward. Cou...
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其他好文 时间:
2014-06-08 15:32:45
阅读次数:
245
extjs4 tree check 级联选择 实现效果:关键代码:
function changeAllNode(node, isCheck) {
allChild(node, isCheck);
allParent(node, isCheck);
function allChild(nodec, isCheckc) {
var chileNodes = n...
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Web程序 时间:
2014-06-08 15:25:20
阅读次数:
222
1. 递归解法
2. 非递归解法(空间复杂度O(n)和O(1))...
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其他好文 时间:
2014-06-08 10:47:37
阅读次数:
139
题目
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
方法
使用DFS对树进行遍...
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其他好文 时间:
2014-06-08 10:26:33
阅读次数:
207
题目
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20...
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其他好文 时间:
2014-06-08 09:23:34
阅读次数:
230
题目:convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
解题思路:这个是个纯粹找规律的题,其他没啥特殊的。下面的例子nRows=4;
找规律按照数组小标开始,寻找下标出现的规律,
1. 第一行和最后一行相邻元素下标之差为 2*nRows-2;
2. 除过第一行和最后一行,其余行要多一个元素,该元素出现的下标和行号有关,比如5 = 1 + 6 - 2,可以总结出规律为 j + 2*nRows-2 - 2*i;
关于 i 和 j 看以看下面...
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其他好文 时间:
2014-06-08 09:11:57
阅读次数:
230