236. Lowest Common Ancestor of a Binary Tree

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.

一刷
题解：这里跟236不一样的是，它不再限定在binary search tree, 而是任意一个binary tree. 然后DFS左右子树去寻找node p和q

/**
 * 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){
           return right;
        }
         
        else{
            if(right == null) return left;
            else return root;
        }
    }
}

二刷
同上，其实会有很多的重复运算，可以考虑用map来表示在这个node的左还是右子树找到[p,q]

/**
 * 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){//not found
            return right;
        }
        if(right == null) return left;
        return root;
    }
}

三刷
用in-order来给每个点标上记号, 在左子树的index都小于root, 右子树的index都大于root

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int index = 0;
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        //in-order traverse
        Map map = new HashMap<>();
        inorder(root, map);
        TreeNode node = root;
        int b = map.get(p), c = map.get(q);
        while(node!=null){
            int a = map.get(node);
            if(a == b || a == c) return node;
            else if(ba && c>a) node = node.right;
            else return node;
        }
        return node;
        
    }
    
    private void inorder(TreeNode root, Map map){
        if(root == null) return;
        inorder(root.left, map);
        map.put(root, index++);
        inorder(root.right, map);
    }
}