236. Lowest Common Ancestor of a Binary Tree最长公共祖先

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______       /              \    ___5__          ___1__   /      \        /      \   6      _2       0       8         /  \         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

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 (E) Lowest Common Ancestor of a Binary Search Tree

题目大意:

给一个二叉树，指定其中的两个节点，寻找他们的最近公共祖先。

最近公共祖先是两个节点的公共的祖先节点且具有最大深度，且允许节点是自己本身的父节点。

如5，1的公共祖先节点是3；5，4的公共祖先节点是5.

代码如下:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public 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;        }else if(left != null) {            return left;        }else {            return right;        }            }}