LeetCode 236. Lowest Common Ancestor of a Binary Tree

[Woodyiiiiiii]

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]

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 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.

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实际上就是树形DP的问题。后序遍历，从左子树和右子树收集信息，到当前根节点处理。

处理信息分为几种情况：

1. 左子树和右子树不为空，则该根节点就是最近公共祖先；
2. 当前根节点为题目所给节点，左子树或右子树信息也返回的是所给节点，那么该根节点就是祖先；3. 左子树或右子树信息返回的是所给节点，而根节点不是，则返回所给节点；
3. 当前根节点为题目所给节点，左子树或右子树信息返回空， 则返回所给节点；
4. 其余情况为空；

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

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参考资料:
LeetCode原题