From b5418913280d2abb580ec35d44c906e25cd221f6 Mon Sep 17 00:00:00 2001 From: CyC2018 <1029579233@qq.com> Date: Thu, 21 Jun 2018 10:34:04 +0800 Subject: [PATCH] auto commit --- notes/Leetcode 题解.md | 237 +++++++++++++++++++++-------------------- 1 file changed, 121 insertions(+), 116 deletions(-) diff --git a/notes/Leetcode 题解.md b/notes/Leetcode 题解.md index e79ad01b..733d5851 100644 --- a/notes/Leetcode 题解.md +++ b/notes/Leetcode 题解.md @@ -5325,6 +5325,70 @@ public int maxDepth(TreeNode root) { } ``` +**平衡树** + +[110. Balanced Binary Tree (Easy)](https://leetcode.com/problems/balanced-binary-tree/description/) + +```html + 3 + / \ + 9 20 + / \ + 15 7 +``` + +平衡树左右子树高度差都小于等于 1 + +```java +private boolean result = true; + +public boolean isBalanced(TreeNode root) { + maxDepth(root); + return result; +} + +public int maxDepth(TreeNode root) { + if (root == null) return 0; + int l = maxDepth(root.left); + int r = maxDepth(root.right); + if (Math.abs(l - r) > 1) result = false; + return 1 + Math.max(l, r); +} +``` + +**两节点的最长路径** + +[543. Diameter of Binary Tree (Easy)](https://leetcode.com/problems/diameter-of-binary-tree/description/) + +```html +Input: + + 1 + / \ + 2 3 + / \ + 4 5 + +Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3]. +``` + +```java +private int max = 0; + +public int diameterOfBinaryTree(TreeNode root) { + depth(root); + return max; +} + +private int depth(TreeNode root) { + if (root == null) return 0; + int leftDepth = depth(root.left); + int rightDepth = depth(root.right); + max = Math.max(max, leftDepth + rightDepth); + return Math.max(leftDepth, rightDepth) + 1; +} +``` + **翻转树** [226. Invert Binary Tree (Easy)](https://leetcode.com/problems/invert-binary-tree/description/) @@ -5332,7 +5396,7 @@ public int maxDepth(TreeNode root) { ```java public TreeNode invertTree(TreeNode root) { if (root == null) return null; - TreeNode left = root.left; // 后面的操作会改变 left 指针，因此先保存下来 + TreeNode left = root.left; // 后面的操作会改变 left 指针，因此先保存下来 root.left = invertTree(root.right); root.right = invertTree(left); return root; @@ -5424,12 +5488,12 @@ Return 3. The paths that sum to 8 are: ```java public int pathSum(TreeNode root, int sum) { - if(root == null) return 0; + if (root == null) return 0; int ret = pathSumStartWithRoot(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum); return ret; } -private int pathSumStartWithRoot(TreeNode root, int sum){ +private int pathSumStartWithRoot(TreeNode root, int sum) { if (root == null) return 0; int ret = 0; if (root.val == sum) ret++; @@ -5449,10 +5513,12 @@ Given tree s: 4 5 / \ 1 2 + Given tree t: 4 / \ 1 2 + Return true, because t has the same structure and node values with a subtree of s. Given tree s: @@ -5464,10 +5530,12 @@ Given tree s: 1 2 / 0 + Given tree t: 4 / \ 1 2 + Return false. ``` @@ -5503,7 +5571,7 @@ public boolean isSymmetric(TreeNode root) { return isSymmetric(root.left, root.right); } -private boolean isSymmetric(TreeNode t1, TreeNode t2){ +private boolean isSymmetric(TreeNode t1, TreeNode t2) { if (t1 == null && t2 == null) return true; if (t1 == null || t2 == null) return false; if (t1.val != t2.val) return false; @@ -5511,37 +5579,6 @@ private boolean isSymmetric(TreeNode t1, TreeNode t2){ } ``` -**平衡树** - -[110. Balanced Binary Tree (Easy)](https://leetcode.com/problems/balanced-binary-tree/description/) - -```html - 3 - / \ - 9 20 - / \ - 15 7 -``` - -平衡树左右子树高度差都小于等于 1 - -```java -private boolean result = true; - -public boolean isBalanced(TreeNode root) { - maxDepth(root); - return result; -} - -public int maxDepth(TreeNode root) { - if (root == null) return 0; - int l = maxDepth(root.left); - int r = maxDepth(root.right); - if (Math.abs(l - r) > 1) result = false; - return 1 + Math.max(l, r); -} -``` - **最小路径** [111. Minimum Depth of Binary Tree (Easy)](https://leetcode.com/problems/minimum-depth-of-binary-tree/description/) @@ -5591,6 +5628,7 @@ private boolean isLeaf(TreeNode node){ ```html Input: + 3 / \ 0 4 @@ -5603,6 +5641,7 @@ Input: R = 3 Output: + 3 / 2 @@ -5644,81 +5683,19 @@ private TreeNode toBST(int[] nums, int sIdx, int eIdx){ } ``` -**两节点的最长路径** - -[543. Diameter of Binary Tree (Easy)](https://leetcode.com/problems/diameter-of-binary-tree/description/) - -```html -Input: - 1 - / \ - 2 3 - / \ - 4 5 - -Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3]. -``` - -```java -private int max = 0; - -public int diameterOfBinaryTree(TreeNode root) { - depth(root); - return max; -} - -private int depth(TreeNode root) { - if (root == null) return 0; - int leftDepth = depth(root.left); - int rightDepth = depth(root.right); - max = Math.max(max, leftDepth + rightDepth); - return Math.max(leftDepth, rightDepth) + 1; -} -``` - -**找出二叉树中第二小的节点** - -[671. Second Minimum Node In a Binary Tree (Easy)](https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/description/) - -```html -Input: - 2 - / \ - 2 5 - / \ - 5 7 - -Output: 5 -``` - -一个节点要么具有 0 个或 2 个子节点，如果有子节点，那么根节点是最小的节点。 - -```java -public int findSecondMinimumValue(TreeNode root) { - if (root == null) return -1; - if (root.left == null && root.right == null) return -1; - int leftVal = root.left.val; - int rightVal = root.right.val; - if (leftVal == root.val) leftVal = findSecondMinimumValue(root.left); - if (rightVal == root.val) rightVal = findSecondMinimumValue(root.right); - if (leftVal != -1 && rightVal != -1) return Math.min(leftVal, rightVal); - if (leftVal != -1) return leftVal; - return rightVal; -} -``` - **二叉查找树的最近公共祖先** [235. Lowest Common Ancestor of a Binary Search Tree (Easy)](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/) ```html _______6______ - / \ - ___2__ ___8__ - / \ / \ - 0 _4 7 9 - / \ - 3 5 + / \ + ___2__ ___8__ + / \ / \ +0 4 7 9 + / \ + 3 5 + For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition. ``` @@ -5736,12 +5713,13 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { ```html _______3______ - / \ - ___5__ ___1__ - / \ / \ - 6 _2 0 8 - / \ - 7 4 + / \ + ___5__ ___1__ + / \ / \ +6 2 0 8 + / \ + 7 4 + For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition. ``` @@ -5804,17 +5782,44 @@ Maximum amount of money the thief can rob = 3 + 3 + 1 = 7. public int rob(TreeNode root) { if (root == null) return 0; int val1 = root.val; - if (root.left != null) { - val1 += rob(root.left.left) + rob(root.left.right); - } - if (root.right != null) { - val1 += rob(root.right.left) + rob(root.right.right); - } + if (root.left != null) val1 += rob(root.left.left) + rob(root.left.right); + if (root.right != null) val1 += rob(root.right.left) + rob(root.right.right); int val2 = rob(root.left) + rob(root.right); return Math.max(val1, val2); } ``` +**找出二叉树中第二小的节点** + +[671. Second Minimum Node In a Binary Tree (Easy)](https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/description/) + +```html +Input: + 2 + / \ + 2 5 + / \ + 5 7 + +Output: 5 +``` + +一个节点要么具有 0 个或 2 个子节点，如果有子节点，那么根节点是最小的节点。 + +```java +public int findSecondMinimumValue(TreeNode root) { + if (root == null) return -1; + if (root.left == null && root.right == null) return -1; + int leftVal = root.left.val; + int rightVal = root.right.val; + if (leftVal == root.val) leftVal = findSecondMinimumValue(root.left); + if (rightVal == root.val) rightVal = findSecondMinimumValue(root.right); + if (leftVal != -1 && rightVal != -1) return Math.min(leftVal, rightVal); + if (leftVal != -1) return leftVal; + return rightVal; +} +``` + ### 层次遍历 使用 BFS 进行层次遍历。不需要使用两个队列来分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。