Leetcode解题: Lowest Common Ancestor of a Binary Tree (236)

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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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

    _______3______
   /              \
___5__          ___1__

/ \ / \
6 _2 0 8
/ \
7 4
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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解法1： O（N）

这题和Lowest Common Ancestor of a Binary Search Tree (235)的不同之处在于这里的条件比较松，只知道是一个binary tree。所以当root不是其中一个node的时候，需要同时计算left和right是否有存在的LCA。
那么递归的返回条件就是只要找到root和其中之一相等的时候，就返回当前的node。
这样就会出现种情况，
root 不是两个node的任何一个，

  1. LCA可以存在于left tree中
  2. 也可以存在于right tree中
  3. 也可以是当前的root（因为左右各含一个node）。
    对于1，2两种情况，其中一个tree会返回空值
    对于第三种情况，left和right的值都不同，以此就可以区分而得出我们想要的结果。

C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        
        if (root == NULL || root == p || root == q) {
            return root;
        }
        
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        
        if (left == NULL) {
            return right;
        }
        
        if (right == NULL) {
            return left;
        }
        
        return root;
    }
};

Java

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) {
            return root;
        }
        
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left == null) {
            return right;
        }
        
        if (right == null) {
            return left;
        }
        
        return root;
    }
}