[LeetCode] 236. Lowest Common Ancestor of a Binary Tree 二叉树的最近公共祖先

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

Example 1:

5
1
3.

Example 2:

5
4
5

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the binary tree.

与 235. Lowest Common Ancestor of a Binary Search Tree 类似，这道题是普通二叉树，不是二叉搜索树，所以不能利用二叉搜索树的性质，所以只能在二叉树中来搜索p和q，然后从路径中找到最近公共祖先。

解法：迭代

Java:

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

Python:

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class Solution:     # @param {TreeNode} root     # @param {TreeNode} p     # @param {TreeNode} q     # @return {TreeNode}     def lowestCommonAncestor(self, root, p, q):         if root in (None, p, q):             return root           left, right = [self.lowestCommonAncestor(child, p, q) \                          for child in (root.left, root.right)]         # 1. If the current subtree contains both p and q,         #    return their LCA.         # 2. If only one of them is in that subtree,         #    return that one of them.         # 3. If neither of them is in that subtree,         #    return the node of that subtree.         return root if left and right else left or right

C++：Recursion

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class Solution { public:     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {        if (!root || p == root || q == root) return root;        TreeNode *left = lowestCommonAncestor(root->left, p, q);        TreeNode *right = lowestCommonAncestor(root->right, p , q);        if (left && right) return root;        return left ? left : right;     } };

C++:

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class Solution { public:     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {         if (!root || root == p || root == q) {             return root;         }         TreeNode *left = lowestCommonAncestor(root->left, p, q);         TreeNode *right = lowestCommonAncestor(root->right, p, q);          // 1. If the current subtree contains both p and q,          //    return their LCA.          // 2. If only one of them is in that subtree,          //    return that one of them.          // 3. If neither of them is in that subtree,          //    return the node of that subtree.         return left ? (right ? root : left) : right;     } };

　　

类似题目：　　

[LeetCode] 235. Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先

 

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