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  • leetcode--Path Sum

    Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

    For example:
    Given the below binary tree and sum = 22,

                  5
                 / 
                4   8
               /   / 
              11  13  4
             /        
            7    2      1
    

    return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

    dfs method:

    public class Solution {
        public boolean hasPathSum(TreeNode root, int sum) {
            if(root == null) return false; 
            return dfs(root, 0, sum);
        }
            
        private boolean dfs (TreeNode node, int accu, int sum) {
            accu += node.val;
            if(node.left == null && node.right == null && accu == sum)
                return true;
            else {
                if(node.left != null) {
                    if(dfs(node.left, accu, sum))
                        return true;
                } 
                            
                if(node.right != null) {
                    if(dfs(node.right, accu, sum))
                        return true;
                }
            }
            accu -= node.val;
            return false;
        }
    }
    

      

    another code:

    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        /**
    	 * This problem can be solved either by dfs or bfs.<br>
    	 * @param root
    	 * @param sum
    	 * @return
    	 */
        public boolean hasPathSum(TreeNode root, int sum) {
            boolean exists = false;
    		if(root != null){
    			Map<TreeNode, Integer> sumAtNode = new HashMap<TreeNode, Integer>();
    			sumAtNode.put(root, root.val);
    			Deque<TreeNode> nodes = new LinkedList<TreeNode>();
    			nodes.add(root);
    			while(nodes.peek()!= null){
    				TreeNode aNode = nodes.poll();
    				if(aNode.left == null && aNode.right == null){
    					if(sumAtNode.get(aNode) == sum){
    						exists = true;
    						break;
    					}
    				}
    				else{
    					int pareValue = sumAtNode.get(aNode);
    					if(aNode.left != null){
    						sumAtNode.put(aNode.left, aNode.left.val + pareValue);
    						nodes.add(aNode.left);
    					}
    					if(aNode.right != null){
    						sumAtNode.put(aNode.right, aNode.right.val + pareValue);
    						nodes.add(aNode.right);
    					}
    				}
    			}
    		}
    		return exists;   
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/averillzheng/p/3773952.html
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