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 nodes5
and1
is3.
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 nodes5
and4
is5
, 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.
题目大意:
寻找两个子节点的最低公共子节点
解法:
这一题没做出来,看了网上的解法。递归的寻找子节点p和子节点q是否在树中出现过,如果说子节点在左右子树哪一个子树中出现过,就返回该子树值,如果在两个子树都出现过,说明公共的最低节点是父节点,如果只在一个子树中出现过,那么公共节点就是左右子树的父节点。
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?left:right; } }