Binary Tree - Lowest Common Ancestor 題型總結

Lowest Common Ancestor of a Binary Search Tree

 Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

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 allowa node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

思路：這題跟 Lowest Common Ancestor of Binary Tree 一模一樣。思路：就是找p的節點在不在左支，或者右支，找到各自左右節點，然後進行比較，如果兩者不一樣，說明當前的root就是lowest 父節點，如果左邊爲空，那就都在右邊，返回右邊的即可，如果右邊爲空，那就都在左邊，返回左邊的即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null) {
            return null;
        }
        if(p == null || q == null) {
            return root;
        }
        TreeNode node = root;
        while(node != null) {
            if(node.val < p.val && node.val < q.val) {
                node = node.right;
            } else if(node.val > p.val && node.val > q.val) {
                node = node.left;
            } else {
                break;
            }
        }
        return node;
    }
}

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: root of the tree
     * @param p: the node p
     * @param q: the node q
     * @return: find the LCA of p and q
     */
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null) {
            return null;
        }
        
        if(p == root || q == root) {
            return root;
        }
        
        if(p.val < root.val && q.val < root.val) {
            return lowestCommonAncestor(root.left, p, q);
        }
        
        if(p.val > root.val && q.val > root.val) {
            return lowestCommonAncestor(root.right, p, q);
        }
        
        return root;
    }
}

 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 allowa 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.

思路：lowestCommonAncestor 的定義就是找到的LCA；

如果兩者都不爲空，說明當前的root就是lowest 父節點，如果左邊爲空，那就都在右邊，返回右邊的即可，如果右邊爲空，那就都在左邊，返回左邊的即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root == q) {
            return root;
        }
        TreeNode leftnode = lowestCommonAncestor(root.left, p, q);
        TreeNode rightnode = lowestCommonAncestor(root.right, p, q);
        if(leftnode == null) {
            return rightnode;
        } else if(rightnode == null) {
            return leftnode;
        } else {
            return root;
        }
    }
}

Given a rooted binary tree, return the lowest common ancestor of its deepest leaves.

Recall that:

• The node of a binary tree is a leaf if and only if it has no children
• The depth of the root of the tree is 0, and if the depth of a node is d, the depth of each of its children is d+1.
• The lowest common ancestor of a set S of nodes is the node A with the largest depth such that every node in S is in the subtree with root A.

Example 1:

Input: root = [1,2,3]
Output: [1,2,3]
Explanation: 
The deepest leaves are the nodes with values 2 and 3.
The lowest common ancestor of these leaves is the node with value 1.
The answer returned is a TreeNode object (not an array) with serialization "[1,2,3]".

Example 2:

Input: root = [1,2,3,4]
Output: [4]

Example 3:

Input: root = [1,2,3,4,5]
Output: [2,4,5]

思路：題目要求求deepest leaf的LCA，我們首先需要tree depth的信息(注意不是node depth, 也可以理解爲deepest leaf depth 也就是tree depth信息)，然後跟LCA一樣，需要返回node信息，那麼我們就需要resultType作爲返回值；findLCA 表示當前枝，找到的LCA和它所能找到的deepest leaf 的depth；如果左右depth相等，證明當前node就是LCA；並返回leftnode的depth也就是deepest node的depth；

注意這裏有兩個表示：一個是method的depth代表node的depth，另外一個returnType裏面的depth代表找到的node的 deepest depth；

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private class ReturnType {
        public TreeNode node;
        public int depth;
        public ReturnType(TreeNode node, int depth) {
            this.node = node;
            this.depth = depth;
        }
    }
    
    public TreeNode lcaDeepestLeaves(TreeNode root) {
        if(root == null) {
            return null;
        }
        ReturnType n = findLCA(root, 0);
        return n.node;
    }
    
    private ReturnType findLCA(TreeNode root, int depth) {
        if(root == null) {
            return new ReturnType(null, depth);
        }
        ReturnType leftnode = findLCA(root.left, depth + 1);
        ReturnType rightnode = findLCA(root.right, depth + 1);
        if(leftnode.depth == rightnode.depth) {
            return new ReturnType(root, leftnode.depth);
        }
        if(leftnode.depth > rightnode.depth) {
            return leftnode;
        } else {
            return rightnode;
        }
    }
}