236. Lowest Common Ancestor of a Binary Tree

題目：

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.

鏈接： http://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

題解：

普通二叉樹求公共祖先。看過<劍指Offer>以后知道這道題應該形成一系列問題。比如是不是二叉樹，是不是BST。假如是BST的話我們可以用上題的方法，二分搜索。有沒有指向父節點的link，假如有指向父節點的link我們就可以用intersection of two lists的方法找到兩個linked list相交的地方。 對這道題目，我們使用后續遍歷來做：

1. 定義兩個輔助節點，使用后續遍歷來遍歷整個樹
2. 當root的值等於p或者q時，找到一個符合條件的節點，返回這個root
3. 先遍歷左子樹
4. 再遍歷右子樹
5. 當left，right均找到時返回此root
6. 只找到left時返回left
7. 只找到right時返回right
8. 否則返回null

Time Complexity - O(n)， Space Complexity - O(n)

public class Solution {
    public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) 
            return null;
        if (root == p || root == q) 
            return root;
        
        TreeNode left = lowestCommonAncestor(root.left, p, q);      // Post order traveral
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        
        if (left != null && right != null)          // p and q in two subtrees
            return root;
        else
            return left != null ? left : right; 
    }
}

 

二刷:

Java:

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

 

三刷

還是跟以前一樣的方法，利用遞歸，先遍歷兩個子樹，來查找是否其中含有目標節點p或者q。假如兩節點分別位於root的左右兩側，則root為LCA. 否則，left和right哪個非空，則哪一個為LCA, 這一側含有p和q兩個目標節點。

Java:

/**
 * 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) return null;
        if (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 return (left != null) ? left : right;
    }
}

 

 

 

Reference: