【leetcode】236. Lowest Common Ancestor of a Binary Tree

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本文深入探讨了二叉树中寻找两个节点最低公共祖先(LCA)的算法，通过自底向上的递归方式，详细解释了如何高效地解决这一经典问题。文章提供了具体的实现代码，帮助读者理解并掌握LCA算法的核心思想。

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.

思考：这是一道二叉树搜索最低公共祖先结点的题，按照以往的经验，我们知道，树的问题一般可以采用递归的方式来进行编程。那么，是采用自顶向下还是自底向上呢？由于需要找最低的公共祖先结点，所以应该采用自底向上的递归方式进行编程。而且由题意我们容易知道给出的二叉树数据结点都是唯一的，所以，当我们在左子树找到p结点，而在右子树找到q结点时，我们便可以找到当前结点为p,q结点的最低公共祖先结点（因为是自底向上递归的）这里还需考虑到，如果p（q）为q(p)的祖先结点，那么p(q)就是p,q的祖先结点了。

实现代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    private TreeNode ans;
    
    public Solution(){
        this.ans = null;
    }
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        //TreeNode ans = null;
        traverse(root,p,q);
        return ans;
    }
    
    public boolean traverse(TreeNode currentNode,TreeNode p,TreeNode q){
        if(currentNode == null){
            return false;
        }
        int left = this.traverse(currentNode.left,p,q)? 1:0;
        int right = this.traverse(currentNode.right,p,q)? 1:0;
        int mid = (currentNode==p || currentNode==q)? 1:0;
        if(left + right +mid >= 2){
            ans = currentNode;
        }
        
        return (mid + left + right >0);
    }
}