Question

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).”

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.

Code

/**
 * 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;
        
        List<TreeNode> pathP = new ArrayList<>();
        List<TreeNode> pathQ = new ArrayList<>();
        pathP.add(root);
        pathQ.add(root);
        
        getPath(root, p, pathP);
        getPath(root, q, pathQ);
        
        TreeNode result = null;
        for (int i = 0; i < pathP.size() && i < pathQ.size(); i++) {
            if (pathP.get(i) == pathQ.get(i)) {
                result = pathP.get(i);
            } else {
                break;
            }
        }
        return result;
    }
    
    private boolean getPath(TreeNode root, TreeNode n, List<TreeNode> path) {  
        if(root==n) {  
            return true;  
        }  
          
        if(root.left!=null) {  
            path.add(root.left);  
            if(getPath(root.left, n, path)) return true;  
            path.remove(path.size()-1);  
        }  
          
        if(root.right!=null) {  
            path.add(root.right);  
            if(getPath(root.right, n, path)) return true;  
            path.remove(path.size()-1);  
        }  
          
        return false;  
    }  
}

Solution

遍历所有结点，找到p, q，同时记录下路径。遍历两条路径，最后一个相同的结点即为结果。