第22天 Binary Tree 235、701、450

235. Lowest Common Ancestor of a Binary Search Tree

当我们从上向下去递归遍历，第一次遇到 cur节点是数值在[p, q]区间中，那么cur就是 p和q的最近公共祖先
不需要遍历整棵树，找到结果直接返回 -> 这里是一条边的写法
方法1: Recursion
Time Complexity: O(n). In the worst case, we might be visiting all the nodes of the binary tree.
Space Complexity: O(n). In the worst case, the recursion stack would be N since the height of a skewed binary tree could be N.

class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root == null || p == root || q == root) return root;TreeNode left = lowestCommonAncestor(root.left, p, q);TreeNode right = lowestCommonAncestor(root.right, p, q);if (left != null && right != null) return root;if (left == null) return right;return left;}
}

方法2: Iteration - 比Binary Tree的要简单，利用只走一边的特性

701. Insert into a Binary Search Tree

方法1: Recursion
Time Complexity: O(H), where H is a tree height. That results in O(log⁡N) in the average case and O(N) in the worst case.
Space Complexity: O(H)

class Solution {public TreeNode insertIntoBST(TreeNode root, int val) {if (root == null) return new TreeNode(val);if (val > root.val) root.right = insertIntoBST(root.right, val);else root.left = insertIntoBST(root.left, val);return root;}
}// 没有必要特别考虑leafclass Solution {public TreeNode insertIntoBST(TreeNode root, int val) {if (root == null) return new TreeNode(val);if (root.left == null && root.right == null) {if (root.val < val) root.right = new TreeNode(val);else root.left = new TreeNode(val);return root;}if (val > root.val) root.right = insertIntoBST(root.right, val);else root.left = insertIntoBST(root.left, val);return root;}
}

方法2: Iteration - 注意BST的iteration和BT的inorder traversal iteration完全不一样
利用只走一边的特性

class Solution {public TreeNode insertIntoBST(TreeNode root, int val) {if (root == null) return new TreeNode(val);TreeNode newRoot = root;TreeNode pre = root;while (root != null) {pre = root;if (root.val > val) {root = root.left;} else if (root.val < val) {root = root.right;} }if (pre.val > val) {pre.left = new TreeNode(val);} else {pre.right = new TreeNode(val);}return newRoot;}
}

450. Delete Node in a BST

五种情况：
- 1. 没找到key
- 2. root.val == key && left == null && right == null
- 3. root.val == key && left == null
- 4. root.val == key && right == null
- 5. root.val == key && left != null && right != null
  - 将删除节点的左子树放到删除节点的右子树的最左面节点的左孩子的位置。并返回删除节点右孩子为新的根节点。

class Solution {public TreeNode deleteNode(TreeNode root, int key) {if (root == null) return null;if (root.val == key) {if (root.left == null) return root.right;else if (root.right == null) return root.left;else {TreeNode cur = root.right;while (cur.left != null) {cur = cur.left;}cur.left = root.left;root = root.right;return root;}}if (root.val > key) root.left = deleteNode(root.left, key);else root.right = deleteNode(root.right, key);return root;}
}

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第22天 Binary Tree 235、701、450

235. Lowest Common Ancestor of a Binary Search Tree

701. Insert into a Binary Search Tree

450. Delete Node in a BST

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