[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:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself
             according to the LCA definition.

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:

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:

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

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++:

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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