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236-Lowest-Common-Ancestor-of-a-Binary-Tree

问题描述

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:

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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:

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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:

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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.

解题思路

  • 在递归函数中,首先看当前结点是否为空,若为空则直接返回空,若为p或q中的任意一个,也直接返回当前结点。否则的话就对其左右子结点分别调用递归函数,由于这道题限制了p和q一定都在二叉树中存在,那么如果当前结点不等于p或q,p和q要么分别位于左右子树中,要么同时位于左子树,或者同时位于右子树。

  • 若p和q分别位于左右子树中,那么对左右子结点调用递归函数,会分别返回p和q结点的位置,而当前结点正好就是p和q的最小共同父结点,直接返回当前结点即可,这就是题目中的例子1的情况。

  • 若p和q同时位于左子树,这里有两种情况,一种情况是 left 会返回p和q中较高的那个位置,而 right 会返回空,所以最终返回非空的 left 即可,这就是题目中的例子2的情况。还有一种情况是会返回p和q的最小父结点,就是说当前结点的左子树中的某个结点才是p和q的最小父结点,会被返回。

  • 若p和q同时位于右子树,同样这里有两种情况,一种情况是 right 会返回p和q中较高的那个位置,而 left 会返回空,所以最终返回非空的 right 即可,还有一种情况是会返回p和q的最小父结点,就是说当前结点的右子树中的某个结点才是p和q的最小父结点,会被返回。

  • 代码如下:

  • JavaScript

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var lowestCommonAncestor = function(root, p, q) {
if (!root || root === p || root === q) return root;
var resL = lowestCommonAncestor(root.left, p, q);
var resR = lowestCommonAncestor(root.right, p, q);
return (resL && resR) ? root : (resL || resR);
};
  • 时间复杂度:O(n)
  • 空间复杂度:O(1)