[LeetCode] 236. Lowest Common Ancestor of a Binary Tree

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).”

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.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

二叉树的最近公共祖先。

题目跟235题很接近，唯一的不同是235题是一个二叉搜索树（BST），本题是一个二叉树。235题可以用递归做，但是本题比较难，需要用到后序遍历（postorder），因为无法用BST的性质判断p和q到底是在左子树还是在右子树。注意需要判断一个corner case，如果根节点为空，或者p和q里面有任何一个节点的 val 和根节点的 val 一样的情况，此时需要返回 root。

时间O(n)

空间O(h) - 树的高度

Java实现

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 && right != null) {
            return root;
        }
        return left == null ? right : left;
    }
}

JavaScript实现

/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function (root, p, q) {
    if (root == null || root == p || root == q) return root;
    let left = lowestCommonAncestor(root.left, p, q);
    let right = lowestCommonAncestor(root.right, p, q);
    if (left !== null && right !== null) {
        return root;
    }
    return left == null ? right : left;
};

根据代码，跑一下example 1，p = 5，q = 1。因为是后序遍历所以先从叶子节点开始找，节点7和节点4满足11行（JS）所以会return他们本身；再往上找的时候会找到他们的父节点2，再找到6和2的父节点5。此时5和1满足11行，所以会return根节点3。

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235. Lowest Common Ancestor of a Binary Search Tree

236. Lowest Common Ancestor of a Binary Tree

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