leetcode [236]Lowest Common Ancestor of a Binary Tree-优快云博客

博客围绕在二叉树中寻找两个给定节点的最低公共子节点展开。介绍了最低公共子节点的定义，给出示例。博主未做出题目，参考网上解法，通过递归寻找子节点是否在树中出现，根据出现情况确定最低公共子节点。

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.

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;
    }
}

转载于:https://www.cnblogs.com/xiaobaituyun/p/10833504.html