236. Lowest Common Ancestor of a Binary Tree

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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.

随便给定一个二叉树，求两个节点的最近公共祖先。

普通的树不具有BST的性质，但受到BST最近公共祖先问题路径的启发，同样可以用记录路径的方法来处理。

先对P做一边搜索，搜索的过程【递归地】“标记”节点是否在根节点到P的路径上，以后序的方法遍历能够保证子树的“标记”在根节点之前。

标记完毕路径塞入hashmap以后来处理Q，处理的方法和Q类似，依然是搜索标记一遍，由于后序遍历，子树的标记优先于根节点，搜索遇到的第一个hashmap中存在的节点（公共节点）就是最近的公共祖先。

public class Solution {    HashMap<TreeNode, Integer> hashmap;TreeNode common=null;public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q){if(root==null)return null;if(p==q)return q;hashmap=new HashMap<>();hashmap.put(root, hashmap.size());search(root, p);find(root, q);return common;}private boolean search(TreeNode t,TreeNode target){if(t==null)return false;if(search(t.left, target)||search(t.right, target)||t==target){hashmap.put(t, hashmap.size());return true;}return false;}private boolean find(TreeNode t,TreeNode target){if(t==null)return false;if(find(t.left, target)||find(t.right, target)||t==target){if(common==null&&hashmap.containsKey(t))    common=t;return true;}return false;}}

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https://discuss.leetcode.com/topic/18566/my-java-solution-which-is-easy-to-understand

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;        return left != null ? left : right;    }}

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