update

Author: jasoncao11

81. 二叉树的直径.md

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+***给定一棵二叉树，你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点。如图所示，二叉树的直径为8***
+
+![algo33](./images/algo33.jpg)
+
+```
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def diameterOfBinaryTree(self, root):
+        self.ans = 0
+        def depth(node):
+            # 访问到空节点了，返回0
+            if not node:
+                return 0
+            # 左儿子为根的子树的深度
+            L = depth(node.left)
+            # 右儿子为根的子树的深度
+            R = depth(node.right)
+            # 计算d_node即L+R 并更新ans
+            self.ans = max(self.ans, L + R)
+            # 返回该节点为根的子树的深度
+            return max(L, R) + 1
+
+        depth(root)
+        return self.ans
+```

82. 二叉树的最近公共祖先.md

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+![algo34](./images/algo34.jpg)
+
+```
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def lowestCommonAncestor(self, root, p, q):
+        """
+        :type root: TreeNode
+        :type p: TreeNode
+        :type q: TreeNode
+        :rtype: TreeNode
+        """
+
+        if not root:
+            return None
+        if (p==root) or (q==root):
+            return root
+
+        left = self.lowestCommonAncestor(root.left, p, q)
+        right = self.lowestCommonAncestor(root.right, p, q)
+        if not left:
+            return right
+        if not right:
+            return left
+        if left and right:
+            return root
+```
+
+二叉搜索树的最近公共祖先:
+```
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+        if p.val < root.val and q.val < root.val:
+            return self.lowestCommonAncestor(root.left, p, q)
+        elif p.val > root.val and q.val > root.val:
+            return self.lowestCommonAncestor(root.right, p, q)
+        return root
+```