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CyC2018
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@ -5325,6 +5325,70 @@ public int maxDepth(TreeNode root) {
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}
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```
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**平衡树**
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[110. Balanced Binary Tree (Easy)](https://leetcode.com/problems/balanced-binary-tree/description/)
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```html
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3
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/ \
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9 20
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/ \
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15 7
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```
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平衡树左右子树高度差都小于等于 1
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```java
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private boolean result = true;
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public boolean isBalanced(TreeNode root) {
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maxDepth(root);
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return result;
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}
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public int maxDepth(TreeNode root) {
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if (root == null) return 0;
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int l = maxDepth(root.left);
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int r = maxDepth(root.right);
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if (Math.abs(l - r) > 1) result = false;
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return 1 + Math.max(l, r);
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}
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```
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**两节点的最长路径**
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[543. Diameter of Binary Tree (Easy)](https://leetcode.com/problems/diameter-of-binary-tree/description/)
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```html
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Input:
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1
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/ \
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2 3
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/ \
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4 5
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Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
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```
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```java
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private int max = 0;
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public int diameterOfBinaryTree(TreeNode root) {
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depth(root);
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return max;
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}
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private int depth(TreeNode root) {
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if (root == null) return 0;
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int leftDepth = depth(root.left);
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int rightDepth = depth(root.right);
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max = Math.max(max, leftDepth + rightDepth);
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return Math.max(leftDepth, rightDepth) + 1;
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}
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```
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**翻转树**
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[226. Invert Binary Tree (Easy)](https://leetcode.com/problems/invert-binary-tree/description/)
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@ -5449,10 +5513,12 @@ Given tree s:
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4 5
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/ \
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1 2
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Given tree t:
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4
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/ \
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1 2
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Return true, because t has the same structure and node values with a subtree of s.
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Given tree s:
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@ -5464,10 +5530,12 @@ Given tree s:
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1 2
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/
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0
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Given tree t:
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4
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/ \
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1 2
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Return false.
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```
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@ -5511,37 +5579,6 @@ private boolean isSymmetric(TreeNode t1, TreeNode t2){
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}
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```
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**平衡树**
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[110. Balanced Binary Tree (Easy)](https://leetcode.com/problems/balanced-binary-tree/description/)
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```html
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3
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/ \
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9 20
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/ \
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15 7
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```
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平衡树左右子树高度差都小于等于 1
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```java
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private boolean result = true;
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public boolean isBalanced(TreeNode root) {
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maxDepth(root);
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return result;
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}
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public int maxDepth(TreeNode root) {
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if (root == null) return 0;
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int l = maxDepth(root.left);
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int r = maxDepth(root.right);
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if (Math.abs(l - r) > 1) result = false;
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return 1 + Math.max(l, r);
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}
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```
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**最小路径**
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[111. Minimum Depth of Binary Tree (Easy)](https://leetcode.com/problems/minimum-depth-of-binary-tree/description/)
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@ -5591,6 +5628,7 @@ private boolean isLeaf(TreeNode node){
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```html
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Input:
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3
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/ \
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0 4
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@ -5603,6 +5641,7 @@ Input:
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R = 3
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Output:
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3
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/
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2
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@ -5644,69 +5683,6 @@ private TreeNode toBST(int[] nums, int sIdx, int eIdx){
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}
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```
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**两节点的最长路径**
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[543. Diameter of Binary Tree (Easy)](https://leetcode.com/problems/diameter-of-binary-tree/description/)
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```html
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Input:
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1
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/ \
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2 3
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/ \
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4 5
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Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
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```
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```java
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private int max = 0;
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public int diameterOfBinaryTree(TreeNode root) {
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depth(root);
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return max;
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}
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private int depth(TreeNode root) {
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if (root == null) return 0;
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int leftDepth = depth(root.left);
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int rightDepth = depth(root.right);
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max = Math.max(max, leftDepth + rightDepth);
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return Math.max(leftDepth, rightDepth) + 1;
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}
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```
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**找出二叉树中第二小的节点**
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[671. Second Minimum Node In a Binary Tree (Easy)](https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/description/)
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```html
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Input:
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2
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/ \
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2 5
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/ \
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5 7
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Output: 5
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```
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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。
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```java
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public int findSecondMinimumValue(TreeNode root) {
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if (root == null) return -1;
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if (root.left == null && root.right == null) return -1;
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int leftVal = root.left.val;
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int rightVal = root.right.val;
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if (leftVal == root.val) leftVal = findSecondMinimumValue(root.left);
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if (rightVal == root.val) rightVal = findSecondMinimumValue(root.right);
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if (leftVal != -1 && rightVal != -1) return Math.min(leftVal, rightVal);
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if (leftVal != -1) return leftVal;
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return rightVal;
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}
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```
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**二叉查找树的最近公共祖先**
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[235. Lowest Common Ancestor of a Binary Search Tree (Easy)](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/)
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@ -5716,9 +5692,10 @@ public int findSecondMinimumValue(TreeNode root) {
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/ \
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___2__ ___8__
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/ \ / \
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0 _4 7 9
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0 4 7 9
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/ \
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3 5
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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.
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```
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@ -5739,9 +5716,10 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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/ \
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___5__ ___1__
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/ \ / \
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6 _2 0 8
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6 2 0 8
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/ \
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7 4
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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.
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```
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@ -5804,17 +5782,44 @@ Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
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public int rob(TreeNode root) {
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if (root == null) return 0;
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int val1 = root.val;
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if (root.left != null) {
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val1 += rob(root.left.left) + rob(root.left.right);
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}
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if (root.right != null) {
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val1 += rob(root.right.left) + rob(root.right.right);
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}
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if (root.left != null) val1 += rob(root.left.left) + rob(root.left.right);
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if (root.right != null) val1 += rob(root.right.left) + rob(root.right.right);
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int val2 = rob(root.left) + rob(root.right);
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return Math.max(val1, val2);
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}
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```
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**找出二叉树中第二小的节点**
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[671. Second Minimum Node In a Binary Tree (Easy)](https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/description/)
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```html
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Input:
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2
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/ \
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2 5
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/ \
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5 7
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Output: 5
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```
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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。
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```java
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public int findSecondMinimumValue(TreeNode root) {
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if (root == null) return -1;
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if (root.left == null && root.right == null) return -1;
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int leftVal = root.left.val;
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int rightVal = root.right.val;
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if (leftVal == root.val) leftVal = findSecondMinimumValue(root.left);
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if (rightVal == root.val) rightVal = findSecondMinimumValue(root.right);
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if (leftVal != -1 && rightVal != -1) return Math.min(leftVal, rightVal);
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if (leftVal != -1) return leftVal;
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return rightVal;
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}
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```
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### 层次遍历
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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。
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